RError.com

RError.com Logo RError.com Logo

RError.com Navigation

  • 主页

Mobile menu

Close
  • 主页
  • 系统&网络
    • 热门问题
    • 最新问题
    • 标签
  • Ubuntu
    • 热门问题
    • 最新问题
    • 标签
  • 帮助
主页 / user-312253

Alexander's questions

Martin Hope
Alexander
Asked: 2024-04-27 20:27:11 +0000 UTC

android.database.CursorIndexOutOfBoundsException:请求索引-1,大小为1

  • 3

我想通过 ID 获取有关我的汽车的信息。但我遇到了一些错误。而且我不明白为什么会发生这种事???

主要活动

DbHandler dbHandler = new DbHandler(this);
Car car1 = dbHandler.getCar(1);

数据库处理程序

public Car getCar(int id){
        SQLiteDatabase db = this.getReadableDatabase();
        Cursor cursor = db.query(Util.TABLE_NAME, new String[]{Util.KEY_ID, Util.KEY_NAME,   Util.KEY_PRICE},
                Util.KEY_ID + "=?", new String[]{String.valueOf(id)}, null, null, null, null);


        if (cursor != null){
            cursor.moveToFirst();
        }
        Car car = new Car(cursor.getInt(0), cursor.getString(1), cursor.getDouble(2));
        return car;
    }

车

public class Car {
    private int id;
    private String name;
    private double price;

    public Car(){

    }

    public Car(String name, double price) {
        this.name = name;
        this.price = price;
    }

    public Car(int id, String name, double price) {
        this.id = id;
        this.name = name;
        this.price = price;
    }

    public int getId() {
        return id;
    }

    public void setId(int id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public double getPrice() {
        return price;
    }

    public void setPrice(double price) {
        this.price = price;
    }
}

我写了“allCars”方法,我得到了所有的 id

public List<Car> allCars(){
        SQLiteDatabase db = this.getReadableDatabase();
        List<Car> carList = new ArrayList<>();

        String allCars = "SELECT * FROM " + Util.TABLE_NAME;
        Cursor cursor = db.rawQuery(allCars, null);


        while(cursor.moveToNext()){
            int id = cursor.getInt(0);
            String name = cursor.getString(1);
            Double price = cursor.getDouble(2);

           // Log.i("allCars: ", "ID: " + id + " Name: " + name + " Price: " + price);

            carList.add(new Car(id, name, price));
        }
        cursor.close();
        db.close();

        return carList;
   }

在此输入图像描述

java
  • 1 个回答
  • 36 Views
Martin Hope
Alexander
Asked: 2021-12-05 06:12:33 +0000 UTC

RSA解密问题

  • 0

我正在研究 RSA 算法,总的来说一切正常,但解密存在问题。对于某些数字,例如 9、10、13,消息未正确解码。 在此处输入图像描述 在此处输入图像描述

编码

private void button_GetKey_Click(object sender, EventArgs e)
    {
        int n = p * q; // Модуль, который используется в открытом ключе
        int f = (p - 1) * (q - 1);  // Функция Эйлера

        List<int> simpleNumbers = generateSimpleNumbers(f);
        int exp = getExponent(simpleNumbers, f);
        // На данном этапе у нас есть открытый ключ {exp, n}
        //Далее необхожимо сформировафть личный ключ 
        int d = generatePrivateKey(exp, f); // личный ключ {d, n}

        // Теперь можем приступать к кодированию
        Random rnd = new Random();

        //Получить случайное число
        double x = rnd.Next(1, n);
        textBox5.Text = x.ToString();
        x = Math.Pow(x, exp);
        x = x % n;
        textBox3.Text = x.ToString();

        // После приступаем к дешифровке
        double y = Math.Pow(x, d);
        y = y % n;
        textBox4.Text = y.ToString();
    }

    private bool isSimpleNumber(int n)
    {
        int count = 0;

        for (int i = 2; i <= n; i++)
        {
            if (n % i == 0) count++;
        }

        if (count == 1)
            return true;
        else
            return false;
    }

    private List<int> generateSimpleNumbers(int n)
    {
        int count;
        List<int> numbers = new List<int>();

        for (int i = 2; i < n; i++)
        {
            count = 0;
            for (int j = 2; j <= i; j++)
            {
                if (i % j == 0) count++;
            }

            if (count == 1)
                numbers.Add(i);
        }

        return numbers;
    }

    // Находим экспоненту
    private int getExponent(List<int> numbers, int fi)
    {
        foreach (int el in numbers)
        {
            if (fi % el != 0)
                return el;
        }

        return 0;
    }

    // Находим личный ключ
    private int generatePrivateKey(int e, int fi)
    {
        int d = e;
        int test = 0;
        do
        {
            d += 1;
            test = (d * e) % fi;

        } while (!isSimpleNumber(d) || test != 1);

        return d;
    }

为什么解密在某些时候会被破坏?毕竟,没有违反条件(x < n)。基于此来源

c#
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2021-11-04 04:03:13 +0000 UTC

随机数发生器

  • 1

任务是编写自己的随机数生成器,即排除 Random 类。实现了一些东西,这里是代码:

function getRandonNumber() {
    var now = new Date(); // берем текущую дату и время

    var hours = now.getHours(); // Извлекаем часы 
    var minutes = now.getMinutes(); // Извлекаем минуты 
    var seconds = now.getSeconds(); // Извлекаем секунды клики
    var milliseconds = (hours * 3600000) + (minutes * 60000) + (seconds * 1000) + now.getMilliseconds(); // Получаем миллисекунды 

    
    var result = milliseconds % 100;
    document.getElementById("for-word").innerHTML = "Случайное число: " + result;
}

此方法生成一个从 0 到 99 的随机数。现在我不知道如何使数字在 15 到 99 的范围内。我该怎么做?

javascript
  • 2 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-09-19 16:54:22 +0000 UTC

在 WordPress 之前用纯 PHP 和 MySQL 或框架开发网站的相关性是什么

  • 0

在 WordPress 之前用纯 PHP 和 MySQL 开发网站有什么相关性?我已经熟悉了网络编程的基础知识并且写了几个网站,现在我想扩展我的视野(我们称之为),但我有一个问题,用 php 开发网站有什么相关性,或者即使您在 CMS WoorPress、Joomla 等之前使用 django 框架。为什么值得自己开发,而不是使用现成的解决方案并对其进行一些修补?

我问了一个与我决定从什么开始编写新项目有关的问题,谁有任何优点和缺点。

php
  • 3 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-05-10 00:57:48 +0000 UTC

ASP.NET MVC。jqgrid不显示数据

  • 1

我正在尝试在 JqGrid 中显示数据,我在这里做了类似Metanit的所有操作 ,除了我从数据库中获取数据。问题是网格是空的,尽管列表中有数据。不知道可能是什么问题?

我要显示的课程

 public class AcademicPerformance{
        public int Id { get; set; }
        public int Student_Id { get; set; }
        public int Semester { get; set; }
        public string Name_Sub { get; set; }
        public int Mark { get; set; }
}

标记

@{
    Layout = null;
}

<html>
<head>
    <title>jqGrid</title>
    <script src="~/Scripts/jquery-3.4.1.min.js" type="text/javascript"></script>
    <link href="~/Content/themes/base/jquery.ui.all.css" rel="stylesheet" type="text/css" />
    <link href="~/Content/jquery.jqGrid/ui.jqgrid.css" rel="stylesheet" type="text/css" />
    <script src="~/Scripts/jquery-ui-1.10.0.min.js" type="text/javascript"></script>
    <script src="~/Scripts/jquery.jqGrid.min.js" type="text/javascript"></script>
    <body>
        <h2>jQGrid</h2>
        <table id="jqg"></table>
        <script type="text/javascript">
$(document).ready(function () {
    $("#jqg").jqGrid({
        url: '@Url.Action("GetData", "Account")',
        datatype: "json",
        colNames: ['Id', 'Студент', 'Семестр', 'Дисциплина', 'Оценка'],
        colModel: [
        { name: 'Id', index: 'Id', width: 30, stype: 'text' },
        { name: 'Student_Id', index: 'Student_Id', width: 150, sortable: true },
        { name: 'Semester', index: 'Semester', width: 150, sortable: true },
        { name: 'Name_Sub', index: 'Name_Sub', width: 150, sortable: true},
        { name: 'Mark', index: 'Mark', width: 150, sortable: true }

        ],
        rowNum: 5, // число отображаемых строк
        loadonce:true, // загрузка только один раз
        sortname: 'Id', // сортировка по умолчанию по столбцу Id
        sortorder: "desc", // порядок сортировки
        caption: "Данные"
    });
});
        </script>
    </body>
</html>

方法

public ActionResult Cabinet(){
    AppContext db = new AppContext();
    int id = Convert.ToInt32(Session["id"]);
    var list = db.AcademicPerformances.Where(x => x.Student_Id == id).ToList();

    foreach(var el in list)
    {
        academicPerformances.Add(new AcademicPerformance { Id = el.Id, Student_Id = el.Student_Id, Semester = el.Semester, Name_Sub = el.Name_Sub, Mark = el.Mark});
    }

    return View();
}
public string GetData(){
    return JsonConvert.SerializeObject(academicPerformances);
}

在此处输入图像描述

在此处输入图像描述

c#
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-03-16 15:21:54 +0000 UTC

如何在 GridLayout 中以编程方式设置按钮之间的间距

  • 0

XML

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:app="http://schemas.android.com/apk/res-auto"
    xmlns:tools="http://schemas.android.com/tools"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    tools:context=".MainActivity"
    android:orientation="vertical">

    <Button
        android:layout_width="match_parent"
        android:layout_height="wrap_content"
        android:text="@string/button_random"
        android:layout_marginBottom="@dimen/marginBottomTop_button"
        android:layout_marginTop="@dimen/margin_button"
        android:layout_marginLeft="@dimen/margin_button"
        android:layout_marginRight="@dimen/margin_button"
        android:textSize="@dimen/text_size"
        android:fontFamily="@font/brusdi"/>

    <Button
        android:layout_width="match_parent"
        android:layout_height="wrap_content"
        android:text="@string/button_question"
        android:layout_marginLeft="@dimen/margin_button"
        android:layout_marginRight="@dimen/margin_button"
        android:textSize="@dimen/text_size"
        android:fontFamily="@font/brusdi"/>




    <ScrollView
        android:layout_width="match_parent"
        android:layout_height="match_parent"
        android:layout_marginTop="@dimen/marginBottomTop_button">

        <GridLayout
            android:orientation="horizontal"
            android:id="@+id/tableGrid"
            android:layout_width="wrap_content"
            android:layout_height="wrap_content"
            android:layout_gravity="center">

        </GridLayout>

    </ScrollView>

</LinearLayout>

爪哇

package com.example.examenatorproject;

import androidx.appcompat.app.AppCompatActivity;


import android.os.Bundle;
import android.text.PrecomputedText;
import android.view.View;
import android.view.ViewGroup;
import android.widget.Button;
import android.widget.FrameLayout;
import android.widget.GridLayout;
import android.widget.GridLayout.LayoutParams;
import android.widget.LinearLayout;
import android.widget.Toast;


public class MainActivity extends AppCompatActivity {


    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        GridLayout gridLayout = (GridLayout)findViewById(R.id.tableGrid);
        gridLayout.removeAllViews();

        Button button = null;
        gridLayout.setColumnCount(3);
        gridLayout.setRowCount(5);


        FrameLayout.LayoutParams layoutParams = new FrameLayout.LayoutParams(FrameLayout.LayoutParams.WRAP_CONTENT, FrameLayout
                .LayoutParams.WRAP_CONTENT);
        layoutParams.leftMargin = 45;
        layoutParams.rightMargin = 45;
        layoutParams.topMargin = 45;
        layoutParams.bottomMargin = 45;

        for(int i = 0; i < 3; i++)
        {
            for (int j = 0; j < 5; j++) {
                button = new Button(this);
                button.setId((5 * i) + 1 + j);
                button.setText(String.valueOf((5 * i) + 1 + j) );

                button.setLayoutParams(layoutParams);

                gridLayout.addView(button);

                button.setOnClickListener(new View.OnClickListener() {
                    @Override
                    public void onClick(View v) {

                        int id = v.getId();
                        Toast.makeText(MainActivity.this, "" + id, Toast.LENGTH_SHORT).show();

                    }});
            }
        }
    }
}

问题是按钮之间的距离没有增加,如何解决?

java
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-03-12 02:44:01 +0000 UTC

XML如何将新节点添加到特定位置

  • 0

XML文件结构:

<?xml version="1.0" encoding="utf-8"?>
<DOCTORS>
    <DOCTOR>
        <surname>Никитина</surname>
        <name>Авготья</name>
        <patronymic>Петровна</patronymic>
        <profession>Терапевт</profession>
        <category>1</category>
        <PATIENT>
            <surname>Понамарев</surname>
            <name>Олег</name>
            <patronymic>Игнатьевич</patronymic>
            <date_birth>12.02.1997</date_birth>
            <category>Инвалид</category>
            <NOTES>
                <date_note>01.03.2020</date_note>
                <diagnos>ОРВИ</diagnos>
                <price>560</price>
            </NOTES>
        </PATIENT>
        <PATIENT>
            <surname>Николаев</surname>
            <name>Георгий</name>
            <patronymic>Николаевич</patronymic>
            <date_birth>17.11.1985</date_birth>
        </PATIENT>
    </DOCTOR>
    <DOCTOR>
        <surname>Романовна</surname>
        <name>Светлана</name>
        <patronymic>Николаевна</patronymic>
        <profession>Дерматолог</profession>
        <category>6</category>
    </DOCTOR>
</DOCTORS>

例如,我想将患者添加到 Dr. 'Romanovna Svetlana Nikolaevna' 现在我添加如下:

XDocument doc = new XDocument(new XElement("DOCTORS"));
// ...
doc.Element("DOCTORS").Element("DOCTOR").Element("DOCTOR").Add(patient);

问题:

有没有其他方法可以添加?特别是,我想摆脱重复的单词.Element("DOCTOR").Element("DOCTOR")

c#
  • 2 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-03-10 22:04:19 +0000 UTC

使用嵌套类导出 xml 文件

  • 0

必须将数据从 MS SQL SERVER DB 导出到 xml。

XML结构(导入文件):

<?xml version="1.0" encoding="utf-8"?>
<DOCTORS>
    <DOCTOR>
        <surname>Никитина</surname>
        <name>Нина</name>
        <patronymic>Петровна</patronymic>
        <profession>Терапевт</profession>
        <category>1</category>
        <PATIENT>
            <surname>Понамарев</surname>
            <name>Олег</name>
            <patronymic>Игнатьевич</patronymic>
            <date_birth>12.02.1997</date_birth>
            <category>Инвалид</category>
            <NOTES>
                <date_note>01.03.2020</date_note>
                <diagnos>ОРВИ</diagnos>
                <price>560</price>
            </NOTES>
        </PATIENT>
        <PATIENT>
            <surname>Николаев</surname>
            <name>Георгий</name>
            <patronymic>Николаевич</patronymic>
            <date_birth>17.11.1985</date_birth>
        </PATIENT>
    </DOCTOR>
    <DOCTOR>
        <surname>Романовна</surname>
        <name>Светлана</name>
        <patronymic>Николаевна</patronymic>
        <profession>Дерматолог</profession>
        <category>6</category>
    </DOCTOR>
</DOCTORS>

医生班:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Xml.Serialization;

namespace WindowsFormsApp1
{
    [XmlRoot(ElementName = "DOCTORS")]
    public class DOCTORS
    {
        [XmlElement(ElementName = "DOCTOR")]
        public List<DoctorXML> DOCTOR { get; set; }
    }
}

DoctorXML 类:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Xml.Serialization;


namespace WindowsFormsApp1
{
    [XmlRoot(ElementName = "DOCTOR")]
    public class DoctorXML
    {
        [XmlElement(ElementName = "surname")]
        public string Surname { get; set; }
        [XmlElement(ElementName = "name")]
        public string Name { get; set; }
        [XmlElement(ElementName = "patronymic")]
        public string Patronymic { get; set; }
        [XmlElement(ElementName = "profession")]
        public string Profession { get; set; }
        [XmlElement(ElementName = "category")]
        public string Category { get; set; }
        [XmlElement(ElementName = "PATIENT")]
        public List<PATIENTXML> PATIENT { get; set; }
    }
}

PatientXML 类:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Xml.Serialization;


namespace WindowsFormsApp1
{
    [XmlRoot(ElementName = "PATIENT")]
    public class PATIENTXML
    {
        [XmlElement(ElementName = "surname")]
        public string Surname { get; set; }
        [XmlElement(ElementName = "name")]
        public string Name { get; set; }
        [XmlElement(ElementName = "patronymic")]
        public string Patronymic { get; set; }
        [XmlElement(ElementName = "date_birth")]
        public string Date_birth { get; set; }
        [XmlElement(ElementName = "category")]
        public string Category { get; set; }
        [XmlElement(ElementName = "NOTES")]
        public NOTESXML NOTES { get; set; }

    }
}

NotesXML 类:

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Xml.Serialization;

namespace WindowsFormsApp1
{
    [XmlRoot(ElementName = "NOTES")]
    public class NOTESXML
    {
        [XmlElement(ElementName = "date_note")]
        public string Date_note { get; set; }
        [XmlElement(ElementName = "diagnos")]
        public string Diagnos { get; set; }
        [XmlElement(ElementName = "price")]
        public string Price { get; set; }

    }

}

XmlSerialization 类:

using System;
using System.Collections.Generic;
using System.IO;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Xml.Serialization;

namespace WindowsFormsApp1
{
    class XmlSerialization<T>
    {
        private Object _obj = typeof(T);

        public Object ReadData(string filePath, Object obj)
        {
            if (obj == null) throw new ArgumentNullException("obj");

            if (File.Exists(filePath))
            {
                using (FileStream fs = new FileStream(filePath, FileMode.Open))
                {
                    XmlSerializer xser = new XmlSerializer(typeof(T));
                    obj = (T)xser.Deserialize(fs);
                    fs.Close();
                }
                return obj;
            }
            throw new Exception("file does not exist");
        }

        public void WriteData(string filePath, Object obj)
        {
            if (File.Exists(filePath)) File.Delete(filePath);
            using (FileStream fs = new FileStream(filePath, FileMode.Create))
            {
                XmlSerializer xser = new XmlSerializer(typeof(T));
                xser.Serialize(fs, obj);
                fs.Close();
            }
        }
    }
}

问题:

我知道您需要使用 WriteData 方法,但是如何从数据库中获取信息,使其具有与文件导入本身相同的结构?毕竟,如果我只写一个请求var q = context.Doctors.Select()...,那么我将收到非结构化的信息。

c#
  • 2 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-03-02 02:58:06 +0000 UTC

解析带有嵌套类的 xml 文件

  • 0

我正在尝试将 xml 文件导入 MS SQL Server,它似乎可以工作,但它不能正常工作。这是xml文件本身:

<?xml version="1.0" encoding="utf-8"?>
<DOCTORS>
    <DOCTOR>
        <surname>Никитина</surname>
        <name>Нина</name>
        <patronymic>Петровна</patronymic>
        <profession>Терапевт</profession>
        <category>1</category>
        <PATIENT>
            <surname>Понамарев</surname>
            <name>Олег</name>
            <patronymic>Игнатьевич</patronymic>
            <date_birth>12.02.1997</date_birth>
            <category>Инвалид</category>
            <NOTES>
                <date_note>01.03.2020</date_note>
                <diagnos>ОРВИ</diagnos>
                <price>560</price>
            </NOTES>
        </PATIENT>
        <PATIENT>
            <surname>Николаев</surname>
            <name>Георгий</name>
            <patronymic>Николаевич</patronymic>
            <date_birth>17.11.1985</date_birth>
        </PATIENT>
    </DOCTOR>
    <DOCTOR>
        <surname>Романовна</surname>
        <name>Светлана</name>
        <patronymic>Николаевна</patronymic>
        <profession>Дерматолог</profession>
        <category>6</category>
    </DOCTOR>
</DOCTORS>

编码:

// импортирую данные доктора

            var items_doctor = doc.Descendants("DOCTOR")
                .Select(x => new
                {
                    Surname = (string)x.Element("surname"),
                    Name = (string)x.Element("name"),
                    Patronymic = (string)x.Element("patronymic"),
                    Profession = (string)x.Element("profession"),
                    Category = (string)x.Element("category")
                });

            foreach (var el in items_doctor)
            {
                using (var context = new MyDbContext())
                {
                    var doctor = new Doctor
                    {
                        Family_name = el.Surname,
                        Name = el.Name,
                        Patronymic = el.Patronymic,
                        Profession = el.Profession,
                        Category = el.Category
                    };


                    context.Doctors.Add(doctor);
                    context.SaveChanges();
                }
            }

                MessageBox.Show("Операция завершена успешно!");

问题是,首先添加的所有医生都没有关于他的病人的信息,但是有必要加载关于一位医生和他的病人的信息,然后我们继续寻找下一位医生,等等。

你能告诉我如何从这样的文件中读取信息:

医生阅读所有信息 => 患者阅读所有信息 => NOTES 阅读所有信息 => 患者阅读所有信息

让我们继续下一个:

医生考虑了所有信息 => ...

c#
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-02-27 14:45:57 +0000 UTC

使用正则表达式交换单词

  • 1

例如,如果有一个输入字符串:alex sam,那么输出应该是:sam alex

 string s = "alex sam";

 string res = Regex.Replace(s, @"([a-z]+)\s+([a-z]+)", "\\2 \\1");

 MessageBox.Show(res);

但所以这些话不会改变地方

c#
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-11-04 23:56:55 +0000 UTC

XML 链接。如何访问下一个节点?

  • 0

XML文件结构:

<?xml version="1.0" encoding="UTF-8" ?>
<employees>
    <employee>
        <fio>Карпов</fio>
        <date>10.09.1978</date>
        <home>11 А Победа</home>
        <phone>87509708898</phone>
    </employee>  
    <list_work>
        <work>
            <name>ЖЭК</name>
            <date_start>10.09.2001</date_start>
            <date_end>10.09.2004</date_end>
        </work>
        <work>
            <name>Маркетинг</name>
            <date_start>11.04.2005</date_start>
            <date_end>10.12.2011</date_end>
        </work>
    </list_work>
    <list_pay>
        <pay>
            <year>2004</year>
            <month>09</month>
            <salary>75000</salary>
        </pay>
        <pay>
            <year>2007</year>
            <month>07</month>
            <salary>250000</salary>
        </pay>
    </list_pay>

    <employee>
    <fio>Глухарь</fio>
    <date>02.12.1981</date>
    <home>12 Б Минусинская</home>
    <phone>87509708881</phone>
<list_work>
    <work>
        <name>Рекламный агент</name>
        <date_start>10.03.2011</date_start>
        <date_end>05.09.2014</date_end>
    </work>
    <work>
        <name>F1</name>
        <date_start>11.04.2016</date_start>
        <date_end>08.10.2018</date_end>
    </work>
</list_work>
<list_pay>
    <pay>
        <year>2012</year>
        <month>06</month>
        <salary>56000</salary>
    </pay>
    <pay>
        <year>2015</year>
        <month>12</month>
        <salary>550000</salary>
    </pay>
</list_pay>
</employee> 
</employees>

我需要选择姓氏、地址和职位并将这些信息加载到 listView 中。

xDoc = XDocument.Load("worker.xml");
            IEnumerable<XElement> workers = xDoc.Elements();

            foreach (XElement worker in workers.Elements())
            {

                listView1.Items.Add(worker.Element("fio").Value);
                listView1.Items[worker.Items.Count - 1].SubItems.Add(worker.Element("home").Value);
                listView1.Items[listView1.Items.Count - 1].SubItems.Add(worker.Element("name").Value);
            }

但是我运行程序时,只显示了姓氏和地址,然后只显示了第一个条目,并且根本没有作品的标题。 在此处输入图像描述

如何让它显示所有条目,即姓氏和地址,以及显示作品的标题?

c#
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-03-26 18:18:16 +0000 UTC

如何获得两个表中的行数之间的差异?

  • 0

如何 MINUS计算表格之间的差异。例如,表 A 有 10 行,B 有 5 结果:5

sql
  • 2 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-03-19 01:48:49 +0000 UTC

甲骨文。请求未完成

  • 1

查询:列出所有收入超过所在部门平均工资的员工

我按部门得出平均工资:

SELECT AVG(salary) 
FROM listworker, listdep
WHERE listworker.department_code = listdep.code_department
GROUP BY (listdep.name_department)

结论:

在此处输入图像描述

接下来我使用子查询:

SELECT worker_surname, name_department, salary
FROM listworker, listdep
WHERE listworker.salary > 
      (SELECT AVG(salary) 
       FROM listworker, listdep
       WHERE listworker.department_code = listdep.code_department
       GROUP BY (listdep.name_department));

但是它返回的值不止一个,如何使它只与部门的平均工资进行比较。

我知道您可以指定一个附加条件,但它不起作用。这里的工作只与农学有关,而与其他部门无关

SELECT worker_surname, name_department, salary
FROM listworker, listdep
WHERE listworker.salary > 
     (SELECT AVG(salary) 
      FROM listworker, listdep
      WHERE listworker.department_code = listdep.code_department 
        AND listdep.name_department = 'Агрономия'
      GROUP BY (listdep.name_department));
база-данных
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-03-14 13:12:22 +0000 UTC

如何加快查询执行时间?

  • 0

我怎样才能加快这个请求:

SELECT id, surname,
CASE
    WHEN ball < 60 THEN 'Низкий'
    WHEN ball BETWEEN 60 AND 80 THEN 'Средний'
    WHEN ball > 80 THEN 'Высокий'
    END BALL
FROM table1

INTERSECT

SELECT id, surname, 
CASE
    WHEN ball < 60 THEN 'Низкий'
    WHEN ball BETWEEN 60 AND 80 THEN 'Средний'
    WHEN ball > 80 THEN 'Высокий'
    END BALL
FROM table2;

使用交叉查询显示其信息包含在两个表中的学生的数据。以及他们的观点的描述。

база-данных
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-02-17 20:09:32 +0000 UTC

SQL 开发人员。如何只留下日期(没有时间)

  • 0

如何从日期中删除时间?以为是这样的:

ALTER SESSION SET NLS_DATE_FORMAT='dd.mm.yyyy';

但是你仍然可以看到时间。 在此处输入图像描述

sql
  • 2 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-12-05 21:48:29 +0000 UTC

如何实现对服务器的请求?

  • -2

我对以下问题感兴趣:从应用程序到服务器的请求是如何实现的?例如:每 10 秒一次。

你能举个例子吗?

android
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-11-21 01:14:04 +0000 UTC

背包问题。蛮力

  • -1

我正在解决背包问题,但我无法进行完整搜索。这是代码:

public class Backpack {
    static int maxWeight = 80;

    Goods[] goods;
    public static void main(String[] args) {
        Goods[] goods = {
                new Goods(15, 30),
                new Goods(30, 90),
                new Goods(50, 100)
        };

        System.out.println("Total: " + findBestRes(goods));
    }

    private static int findBestRes(Goods[] goods){
        int max = 0;


        for (int i = 0; i < goods.length; i++) {
            int temp = goods[i].getValue();
            maxWeight -= goods[i].getWeight();

            for (int j = 0; j < goods.length; j++) {
                if (i != j && maxWeight >= goods[j].getWeight()){
                    temp += goods[j].getValue();
                    maxWeight -= goods[j].getWeight();

                }
            }

            if (temp > max) {
                max = temp;
            }

            maxWeight = 80;
        }

        return max;
    }
}

public class Goods {
    private int weight;
    private int value;

    public Goods(int weight, int value){
        this.weight = weight;
        this.value = value;
    }

    public int getWeight() {
        return weight;
    }

    public int getValue() {
        return value;
    }


}

不要告诉我如何正确实现它或者是否有必要在这里重做所有事情?

java
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-10-15 01:26:11 +0000 UTC

如何画一个方形螺旋?

  • 0

请告诉我如何画一个方形螺旋?一开始我试着用线条来画,但不太成功,也许是这个图的公式?目前我在正方形的帮助下绘制,但结果不是我需要的。

import javax.swing.*;
import java.awt.*;
import java.awt.geom.Line2D;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;


public class Spiral {
    private static JFrame frame;
    private static int x, x2;
    private static int y, y2;
    private static int x1;
    private static int y1;
    private static int n;
    private static int step;
    private static Line2D.Double line = null;

    public Spiral(){
        x = 300;
        y = 200;
        x2 = 400;
        y2 = 200;
        step = 30;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        System.out.println("Enter number: ");
        n = Integer.parseInt(reader.readLine()) ;

        frame = new JFrame("Spiral");
        frame.setSize(600,400);

        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);

        GraphicsComponent gc = new GraphicsComponent();
        frame.add(gc);
        frame.setVisible(true);
    }

    static class GraphicsComponent extends JComponent {
        @Override
        protected void paintComponent(Graphics g) {
            Graphics2D g2 = (Graphics2D) g;
            x1 = 250;
            y1 = 50;

            int size = n * n;
            for (int i = 0; i < n; i++) {
                Rectangle rectangle = new Rectangle(x1, y1, size, size);
                size -= 10;
                x1 += 5;
                y1 += 5;
                g2.draw(rectangle);
            }

            /*
            int dir = 0;
            for (int i = 1; i <= 6; i++) {
                if (dir == 0){
                    line = new Line2D.Double(x, y, x2 ,y2);
                    drawLine(g2);
                    dir++;
                }
                else if(dir == 1){
                    x = x2;
                    y = y;
                    x2 = x2;
                    y2 += STEP;
                    line = new Line2D.Double(x, y, x2 ,y2);
                    drawLine(g2);
                    dir++;
                }
                else if(dir == 2){
                    x = y2 + STEP;
                    y = y2;
                    x2 -= 0;
                    y2 = y2;
                    line = new Line2D.Double(x, y, x2 ,y2);
                    drawLine(g2);
                    dir++;
                }
                else if(dir == 3){
                    x = x;
                    y = y;
                    x2 = x;
                    y2 -= STEP;
                    line = new Line2D.Double(x, y, x2 ,y2);
                    drawLine(g2);
                    dir++;
                }
                else if(dir == 4){
                    x = x;
                    y = y;
                    x2 = x;
                    y2 -= STEP;
                    line = new Line2D.Double(x, y, x2 ,y2);
                    drawLine(g2);
                    dir++;
                }
                else if(dir == 5){
                    x = x;
                    y = y2;
                    x2 += STEP;
                    y2 = y;
                    line = new Line2D.Double(x, y, x2 ,y2);
                    drawLine(g2);
                    dir = 1;
                }
            }
            */
        }
    }

    private static void drawLine(Graphics2D g2){
        g2.setColor(Color.RED);
        g2.fill(line);
        g2.draw(line);
    }
}

方形螺旋

java
  • 1 个回答
  • 10 Views
Martin Hope
Alexander
Asked: 2020-10-14 05:30:06 +0000 UTC

如何从 Firebase 数据库中读取数据?

  • 1

请告诉我如何正确地从 Firebase 读取数据。我根据文档工作,我做了所有这样的事情,但它显示“数据库权限被拒绝”。

package com.example.strike.myapplication;

import android.os.Bundle;
import android.support.annotation.NonNull;
import android.widget.Toast;

import com.google.firebase.database.DataSnapshot;
import com.google.firebase.database.DatabaseError;
import com.google.firebase.database.DatabaseReference;
import com.google.firebase.database.FirebaseDatabase;
import com.google.firebase.database.ValueEventListener;

import stanford.androidlib.SimpleActivity;

public class LoginActivity extends SimpleActivity {
    private FirebaseDatabase database;
    private DatabaseReference myRef;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);

        database = FirebaseDatabase.getInstance();
        myRef = database.getReference("/animal/PLaNGFwshzOhzys2VhOP");

        myRef.addValueEventListener(new ValueEventListener() {
            @Override
            public void onDataChange(@NonNull DataSnapshot dataSnapshot) {
                String value =  dataSnapshot.getValue(String.class);
                Toast.makeText(getApplicationContext(), value, Toast.LENGTH_SHORT).show();
            }

            @Override
            public void onCancelled(@NonNull DatabaseError databaseError) {
                Toast.makeText(getApplicationContext(), "Value is: " + databaseError, Toast.LENGTH_SHORT).show();
            }
        });
    }

}

规则

service cloud.firestore {

    match /databases/{database}/documents {
    match /{document=**} {
       allow read, write;
      }
    }

}
java
  • 1 个回答
  • 10 Views

Sidebar

Stats

  • 问题 10021
  • Answers 30001
  • 最佳答案 8000
  • 用户 6900
  • 常问
  • 回答
  • Marko Smith

    我看不懂措辞

    • 1 个回答
  • Marko Smith

    请求的模块“del”不提供名为“default”的导出

    • 3 个回答
  • Marko Smith

    "!+tab" 在 HTML 的 vs 代码中不起作用

    • 5 个回答
  • Marko Smith

    我正在尝试解决“猜词”的问题。Python

    • 2 个回答
  • Marko Smith

    可以使用哪些命令将当前指针移动到指定的提交而不更改工作目录中的文件?

    • 1 个回答
  • Marko Smith

    Python解析野莓

    • 1 个回答
  • Marko Smith

    问题:“警告:检查最新版本的 pip 时出错。”

    • 2 个回答
  • Marko Smith

    帮助编写一个用值填充变量的循环。解决这个问题

    • 2 个回答
  • Marko Smith

    尽管依赖数组为空,但在渲染上调用了 2 次 useEffect

    • 2 个回答
  • Marko Smith

    数据不通过 Telegram.WebApp.sendData 发送

    • 1 个回答
  • Martin Hope
    Alexandr_TT 2020年新年大赛! 2020-12-20 18:20:21 +0000 UTC
  • Martin Hope
    Alexandr_TT 圣诞树动画 2020-12-23 00:38:08 +0000 UTC
  • Martin Hope
    Air 究竟是什么标识了网站访问者? 2020-11-03 15:49:20 +0000 UTC
  • Martin Hope
    Qwertiy 号码显示 9223372036854775807 2020-07-11 18:16:49 +0000 UTC
  • Martin Hope
    user216109 如何为黑客设下陷阱,或充分击退攻击? 2020-05-10 02:22:52 +0000 UTC
  • Martin Hope
    Qwertiy 并变成3个无穷大 2020-11-06 07:15:57 +0000 UTC
  • Martin Hope
    koks_rs 什么是样板代码? 2020-10-27 15:43:19 +0000 UTC
  • Martin Hope
    Sirop4ik 向 git 提交发布的正确方法是什么? 2020-10-05 00:02:00 +0000 UTC
  • Martin Hope
    faoxis 为什么在这么多示例中函数都称为 foo? 2020-08-15 04:42:49 +0000 UTC
  • Martin Hope
    Pavel Mayorov 如何从事件或回调函数中返回值?或者至少等他们完成。 2020-08-11 16:49:28 +0000 UTC

热门标签

javascript python java php c# c++ html android jquery mysql

Explore

  • 主页
  • 问题
    • 热门问题
    • 最新问题
  • 标签
  • 帮助

Footer

RError.com

关于我们

  • 关于我们
  • 联系我们

Legal Stuff

  • Privacy Policy

帮助

© 2023 RError.com All Rights Reserve   沪ICP备12040472号-5