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Владимир Говоров
Asked:2022-06-14 00:27:36 +0000 UTC2022-06-14 00:27:36 +0000 UTC

如何将一个对象的属性分配给另一个对象。具有相同类型的对象

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有 2 个具有相同类型的对象。一个对象（我们称之为“fulledObj”）最初被填充并遍历键。所有找到的属性都必须在迭代期间写入另一个对象（我们称之为“needfulProps”）。我必须马上说，将一个对象分配给另一个/克隆的解决方案是不合适的。我需要在 needfulProps 对象上覆盖/创建一个属性。

* 非工作部分在代码中标有注释：

                    type responsiveArguments = {
                        heightRow?: string,
                        widthColumn?: string,
                        countColumns?: number,
                        countRows?: number,
                        rowGap?: number
                    }
                    let needfulResponsiveProps: responsiveArguments = {};
                    let oneWidthResponsiveObj: responsiveArguments = {
                        heightRow: "300px",
                        widthColumn: "500px",
                        countRows: 2
                    }
                    for (const keyResponsiveObj of Object.keys(oneWidthResponsiveObj) as (keyof responsiveArguments)[]) {
                        // Вывести в консоль объект с этим свойством я могу
                        console.log(`needfulResponsiveProps[keyResponsiveObj]: `,  needfulResponsiveProps[keyResponsiveObj]);
                        // Присвоить не могу
                        // Тип "string | number | undefined" не может быть назначен для типа "undefined". Тип "string" не может быть назначен для типа "undefined"
                        needfulResponsiveProps[keyResponsiveObj] = oneWidthResponsiveObj[keyResponsiveObj];
                    }

此外，如果对象类型中的所有字段都属于同一类型，则一切正常：

                    type exampleType = { //все поля одного типа(number)
                        num?: number,
                        num2?: number
                    }

                    const filledObj: exampleType = {
                        num: 10,
                        num2: 15
                    }
                    let testConst: exampleType = {};
                    for (const key of Object.keys(filledObj) as (keyof exampleType)[]) {
                        testConst[key] = filledObj[key];
                    }

完整的代码，从中可以清楚为什么简单的对象复制是不合适的：

type responsive = {
            [resolution: number]: responsiveArguments;
        }
        type responsiveArguments = {
            heightRow?: string | undefined;
            widthColumn?: string | undefined;
            countColumns?: number | undefined;
            countRows?: number | undefined;
            rowGap?: number | undefined;
        }        
        /* for test */viewWidthDevice = 1500; // ширина устройства - будет считаться автоматически
        /* for test */responsive = {
            500: {
                countColumns: 500,
                countRows: 2
            },
            800: {
                countRows: 50,
                heightRow: "500px"
            },
            1900: {
                countColumns: 50
            }
        }
        
        let objPropsLessWidth: responsive = {};
        if(responsive) {
            for(let i = 0; i < viewWidthDevice; i++) {
                if(responsive.hasOwnProperty(i)) {
                    objPropsLessWidth[i] = responsive[i];
                };
            }
            console.log("objPropsLessWidth: ", objPropsLessWidth); //ключ 1900 будет убран, ведь ширин

            if(Object.keys(objPropsLessWidth).length > 0) {
                let needfulResponsiveProps: responsiveArguments = {};

                for (const keyOneResponsiveWidth in objPropsLessWidth) { // enumeration all responsive appropriate properties  
                    const oneWidthResponsiveObj: responsiveArguments = objPropsLessWidth[keyOneResponsiveWidth];
                    
                    //надо так:
                    for (const keyResponsiveObj of Object.keys(oneWidthResponsiveObj) as (keyof responsiveArguments)[]) {
                        //в левом операнде ошибка
                        //Тип "string | number | undefined" не может быть назначен для типа "undefined". Тип "string" не может быть назначен для типа "undefined".
                        needfulResponsiveProps[keyResponsiveObj] = oneWidthResponsiveObj[keyResponsiveObj];
                    }

                    //хочу избавиться от этого промежутка кода в сторону цикла, который выше
                    if(oneWidthResponsiveObj.countColumns) needfulResponsiveProps.countColumns = oneWidthResponsiveObj.countColumns;
                    if(oneWidthResponsiveObj.countRows) needfulResponsiveProps.countRows = oneWidthResponsiveObj.countRows;
                    if(oneWidthResponsiveObj.heightRow) needfulResponsiveProps.heightRow = oneWidthResponsiveObj.heightRow;
                    if(oneWidthResponsiveObj.rowGap) needfulResponsiveProps.rowGap = oneWidthResponsiveObj.rowGap;
                    if(oneWidthResponsiveObj.widthColumn) needfulResponsiveProps.widthColumn = oneWidthResponsiveObj.widthColumn;
                }
            }
        }
typescript

1 个回答

  1. Best Answer

    使用循环，让 ts 理解类型将不起作用（至少我不知道方法）：操场

    interface Smth {
  x: string
  y: number
}

declare var a: Smth, b: Smth

for (const key of Object.keys(a) as (keyof Smth)[]) {
  b[key] = a[key] // Type 'string | number' is not assignable to type 'never'.
}

    但您可以使用演员阵容：操场

    interface Smth {
  x: string
  y: number
}

declare var a: Smth, b: Smth

for (const key of Object.keys(a) as (keyof Smth)[]) {
  b[key] = a[key] as never
}

    你也可以使用一个函数来化学，但我不会，因为这是一堆额外的代码：操场

    interface Smth {
  x: string
  y: number
}

declare var a: Smth, b: Smth

function copy<T, K extends keyof T>(dest: T, source: T, key: K) {
  dest[key] = source[key]
}

for (const key of Object.keys(a) as (keyof Smth)[]) {
  copy(b, a, key);
}
    • 1

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