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Михаил Горячев's questions

Martin Hope
Михаил Горячев
Asked: 2023-02-02 20:40:07 +0000 UTC
  • 5

告诉我出了什么事？我通过 ajax 将表单发送给处理程序：

<form id="sendmsg" method="POST" enctype="multipart/form-data">

<!-- фотка -->
    <input type="file" name="images[]" id="files">
    <button class="uk-button uk-button-secondary"><?echo $lang->selectphoto;?></button>
<!-- фотка -->

<textarea id="textarea" class="uk-textarea standart" name="message" maxlength="1500" required></textarea>

<input type="submit" name="send" value="&#10148;" class="uk-button uk-button-secondary send-button" id="sendbtn">

</form>

阿贾克斯：

<script>
    $(document).ready( function() { 
    $("#sendmsg").submit(function(send){
        send.preventDefault();
        $.ajax({
            url: 'updates/sendmsg.php',
            type: 'POST',
            data: {'files': $('#files')[0].files[0],'idtodb':'<?php echo $partnerid ?>', 'message':$('textarea[name="message"]').val()},
            success: function(html){  
                document.getElementById("sendmsg").reset();
                $("#globalresult").html(html);
            }  
        });
    });
    });
</script>

不选择照片，一切正常，但如果你选择照片，控制台会发誓：

jquery.min.js:2 Uncaught (in promise) TypeError: Failed to execute 'arrayBuffer' on 'Blob': Illegal invocation
    at i (jquery.min.js:2:73313)
    at Dt (jquery.min.js:2:73205)
    at Dt (jquery.min.js:2:73228)
    at S.param (jquery.min.js:2:73524)
    at Function.ajax (jquery.min.js:2:76904)
    at HTMLFormElement.<anonymous> (chat?id=23:509:11)
    at HTMLFormElement.dispatch (jquery.min.js:2:43090)
    at v.handle (jquery.min.js:2:41074)
php
Martin Hope
Михаил Горячев
Asked: 2022-09-05 20:48:09 +0000 UTC
  • 1

这个想法是，在每个主要类别的菜单中，如果有产品，则显示其中包含的第一个产品的图像。如果产品存在，则显示，但如果在下一次迭代中，产品的下一个主要类别不存在，则显示上一个的图像，并且应该有一个存根。告诉我，有什么问题？

在此处输入图像描述

//определение категорий
            $sql = "SELECT name,id FROM category WHERE parrent =''";
            if($result = $conn->query($sql)){ 
                foreach($result as $row){ 

                    echo "<div class=\"uk-card uk-flex uk-flex-center uk-flex-middle\">";
                    echo "<div class=\"menu_items\"><div>";

                    $menu_category_name = $row["name"]; 
                    $menu_category_id = $row["id"];

                        //определение картинки главной категории
                        $sql = "SELECT id FROM products WHERE category='$menu_category_id' LIMIT 1";
                        if($result = $conn->query($sql)){
                            foreach($result as $row){

                            $firstproduct_id = $row["id"];

                            if ($firstproduct_id != '') { 

                                $sql = "SELECT image FROM images WHERE product_id ='$firstproduct_id' LIMIT 1";
                                if($result = $conn->query($sql)){ 
                                foreach($result as $row){ 

                                    $categoryimage = $row["image"];

                                }} else { echo "Ошибка: " . $conn->error;}

                            } else {

                                    $categoryimage = "../ui/img/noimage.jpg";
                                    
                            }
    

                            }} else { echo "Ошибка: " . $conn->error;} 

       
                    echo "<ul class=\"uk-nav-default\" uk-nav style=\"padding-left:14px;\"><li><a href=\"category.php?id=$menu_category_id\"><b>$menu_category_name</b></a></li></ul>";

                    //определение максимального количества субкатегорий
                    $sql = "SELECT name,id FROM category WHERE parrent ='$menu_category_id'";
                    if($result = $conn->query($sql)){
                    $rowsCount = $result->num_rows; // количество полученных строк
                    }

                    //определение субкатегорий
                    $sql = "SELECT name,id FROM category WHERE parrent ='$menu_category_id' LIMIT 4";
                    if($result = $conn->query($sql)){ 
                    foreach($result as $row){  

                        $menu_category_name_sub = $row["name"]; 
                        $menu_category_id_sub = $row["id"];

                        echo "<ul class=\"uk-nav-sub\"><li><a href=\"category.php?id=$menu_category_id_sub\"> - $menu_category_name_sub</a></li></ul>";
                        
                    } } else { echo "Ошибка: " . $conn->error;}


                    if ($rowsCount>=5){
                        echo "<ul class=\"uk-nav-sub\"><li><a href=\"category.php?id=$menu_category_id\"> - Смотреть все</a></li></ul>";
                    } 


                    echo "</div><div><img style=\"max-width:150px; object-fit: cover;\" src=\"$categoryimage\"></div></div>";
                    echo "</div>";
     
            } } else { echo "Ошибка: " . $conn->error;}
php
Martin Hope
Михаил Горячев
Asked: 2022-09-05 21:32:49 +0000 UTC
  • 0

如果产品没有图像（即图像表中根本没有产品的行），我正在尝试制作一个 noimage 存根。

$sql = "SELECT * FROM images WHERE product_id='$id' ORDER BY sort ASC LIMIT 5";
if($result = $conn->query($sql)){ foreach($result as $row) { $image = $row['image'];  

  if(mysqli_num_rows($result) == 0) {
    $image = "../ui/img/noimage.jpg";
  } 
...

存根不出现。如果我对所有产品进行检查并显示此 noimage，则一切正常，对于表中有此行的产品，将显示 noimage。

      if(mysqli_num_rows($result)) {
        $image = "../ui/img/noimage.jpg";
      }

如果您只显示echo mysqli_num_rows($result)有图片的产品的行数，而没有图片的产品则不显示。错误在哪里？这可能是设计问题。这是完整的代码：

<div uk-slider>
<div class="uk-position-relative uk-visible-toggle uk-light" tabindex="-1" loading="lazy">
<ul class="uk-slider-items uk-grid uk-grid-match">

<?

$sql = "SELECT * FROM images WHERE product_id='$id' ORDER BY sort ASC LIMIT 5";

if($result = $conn->query($sql)){

if(mysqli_num_rows($result) == 0) {
  $image = "../ui/img/noimage.jpg";

 ?>

  <li>
  <a href=product.php?id=<?echo $id?>><img src="<?echo $image?>" style="object-fit:cover; max-height:400px; min-height:400px; width:100%;"></a>
  </li>

<?

} else if (mysqli_num_rows($result) > 0) {

foreach($result as $row) { $image = $row['image'];  

?>

  <li>
  <a href=product.php?id=<?echo $id?>><img src="<?echo $image?>" style="object-fit:cover; max-height:400px; min-height:400px; width:100%;"></a>
  </li>

<?

}}} else { echo "Ошибка: " . $conn->error;}

?>

</ul>

    <a class="uk-position-center-left uk-position-small uk-hidden-hover" href="#" uk-slidenav-previous uk-slider-item="previous"></a>
    <a class="uk-position-center-right uk-position-small uk-hidden-hover" href="#" uk-slidenav-next uk-slider-item="next"></a>
  
</div>

<ul id = "dotnav-category" class="uk-slider-nav uk-dotnav uk-flex-center"></ul>

</div>
php
Martin Hope
Михаил Горячев
Asked: 2022-08-28 20:23:10 +0000 UTC
  • 0

我将篮子的数据存储在会话中。从卡中将第一个产品添加到购物车时，产品和数量已正确添加。添加下一个产品时，购物车中所有产品的产品数量都会发生变化，包括第一个产品。如何单独更改数量。

处理程序：

// получаем переменные из карточки товара (id и количество)
$item_id = $_POST['id'];
$qty = $_SESSION['cart'][$item_id]['qty'] + $_POST['qty'];

//данные для отображения в корзине
$sql = "SELECT * FROM products WHERE id='$item_id'";
if($result = $conn->query($sql)){ foreach($result as $row){ 
    $product_name = $row["name"]; 
    $product_price = $row["price"]; 
    $product_price_sale = $row["saleprice"]; 

    $sql = "SELECT * FROM images WHERE product_id='$item_id'";
    if($result = $conn->query($sql)){ foreach($result as $row){ $image = $row["image"]; } } else { echo "Ошибка: " . $conn->error;}

} } else { echo "Ошибка: " . $conn->error;}

if ($product_price_sale == ""){
    $finalprice = $product_price . ' р';
} else $finalprice = $product_price_sale . ' р';

//изменение данных сессии
$_SESSION['cart'][$item_id] = ['id' => $item_id, 'image' => $image, 'name' => $product_name, 'qty' => $qty , 'price' => $finalprice];

include "../blocks/cart.php";

?>
php cart
Martin Hope
Михаил Горячев
Asked: 2022-08-03 00:12:47 +0000 UTC
  • 0

如何确保物品分别进入篮子？现在，当添加第一个产品时，一切都很好，但是当添加另一个项目时，位置并没有添加，而是被覆盖了。购物篮中总是有一种产品名称在变化。我将篮子的数据存储在会话中。

处理程序：

$item_id = $_POST['id'];
$qty = $_SESSION['$item_id'][1] + $_POST['qty'];

$sql = "SELECT * FROM products WHERE id='$item_id'";
if($result = $conn->query($sql)){ foreach($result as $row){ $product_name = $row["name"]; } } else { echo "Ошибка: " . $conn->error;}

$_SESSION['$item_id'] = [$product_name, $qty];

结论：

$product_name = $_SESSION['$item_id'][0];
$qty = $_SESSION['$item_id'][1];

<? foreach($_SESSION['$item_id'] AS $value) {?>
<td class="uk-text-nowrap"><?echo $value?></td>
<?}?>
php корзина
Martin Hope
Михаил Горячев
Asked: 2022-07-30 23:17:59 +0000 UTC
  • 1

请告诉我如何正确编写 Ajax 脚本，以便在提交第二个表单时页面不会重新加载。有必要恰好保留 2 个表格。

代码中发生了什么：add_review通过 ajax 提交第一个表单时，我成功提交了第二个表单$('.rating').submit();，但在这种情况下页面会重新加载。是否可以添加脚本以便发送第二个表单但不重新加载页面。

设计：

<form class="rating" method="POST" action="updates/add_review_stars.php" enctype="multipart/form-data">
    <input type="hidden" class="uk-input" value="<?echo $product_id;?>" name="product_id" id="product_id">
    <label>
    <input type="radio" name="stars" value="1" />
    <span class="icon">★</span>
    </label>
    <label>
    <input type="radio" name="stars" value="2" />
    <span class="icon">★</span>
    <span class="icon">★</span>
    </label>
    <label>
    <input type="radio" name="stars" value="3" />
    <span class="icon">★</span>
    <span class="icon">★</span>
    <span class="icon">★</span>   
    </label>
    <label>
    <input type="radio" name="stars" value="4" />
    <span class="icon">★</span>
    <span class="icon">★</span>
    <span class="icon">★</span>
    <span class="icon">★</span>
    </label>
    <label>
    <input type="radio" name="stars" value="5" />
    <span class="icon">★</span>
    <span class="icon">★</span>
    <span class="icon">★</span>
    <span class="icon">★</span>
    <span class="icon">★</span>
    </label>
    </form>

    <form id ="add_review" enctype="multipart/form-data">
    <input type="hidden" class="uk-input" value="<?echo $product_id;?>" name="product_id" id="product_id">
    <input type="hidden" class="uk-input" value="0" name="approved" id="approved">
    </br>
    <label class="uk-form-label" for="form-stacked-text">Имя</label>
    <input type="text" class="uk-input" name="name" id="name" required>
    <label class="uk-form-label" for="form-stacked-text">Отзыв</label>
    <textarea class="uk-textarea" name="review" id="review" required></textarea>
    <div class="uk-margin">
    <input type="submit" name="submit" value="Добавить отзыв" data-dismiss="modal" class="uk-button uk-button-default">
    </div>
    </form>

    <script>
    $("#add_review").submit(function(send){
    send.preventDefault();
    $.ajax({
        url: 'updates/add_review.php',
        type: 'POST',
        data: new FormData(this),
        contentType: false,
        cache: false,
        processData:false,
        success: function(html){  
            $("#addresult_review").html(html);
            $('.rating').submit();
        }  
    });
    });
    </script>
php ajax
Martin Hope
Михаил Горячев
Asked: 2022-06-26 00:09:35 +0000 UTC
  • 0

如何通过单击复选框打开表单中的邮件字段？如果按下，它应该被隐藏。

<form method="POST" action="" enctype="multipart/form-data">
    <input type="text" class="form-control" name="name" required>
    <input name='email' type='checkbox' value="1"> 
    <input type="text" class="form-control" name="mail">
    <input type="submit" name="submit" value="Присоединиться" class="btn btn-success">
</form>
html