Javascript 如何在字段为空时更改数组字段的动态值

以下是我编写的代码：

for (let i = 0; i < result.length; i++) {
    if (result[i].mkp == ' ') {
        //the first time that a empty field happen
        //the array position is filled with "FRANQUIAS"
        result[i].mkp = "FRANQUIAS";
    }
    if (result[i].mkp == ' ') {
        //the second time that a empty field happen
        //the array position is filled with "TELEVENDAS"
        result[i].mkp = "TELEVENDAS";
    }
    if (result[i].mkp == ' ') {
        //the third time that a empty field happen
        //the array position is filled with "OCC"
        result[i].mkp = "OCC";
    }
}

for（设i=0；i

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但是我自己也弄不明白，我怎么能达到把空字段改成这三个字段的目的呢。有人能帮我吗？

保留一个额外的计数器变量，当出现空字段时，它会递增

// variable for keeping track of empty field
let c = 0;

for (let i = 0; i < result.length; i++) {
    // chek value or couter in addition to your condition
    if (c === 0 && result[i].mkp == ' ') {
        //the first time that a empty field happen
        //the array position is filled with "FRANQUIAS"
        result[i].mkp = "FRANQUIAS";
        // increment counter value whenever empty field occurs
        c++;
    }
    else if (c === 1 && result[i].mkp == ' ') {
        //the second time that a empty field happen
        //the array position is filled with "TELEVENDAS"
        result[i].mkp = "TELEVENDAS";
        c++;
    }
    else if (c === 2 && result[i].mkp == ' ') {
        //the third time that a empty field happen
        //the array position is filled with "OCC"
        result[i].mkp = "OCC";
        c++;
    }
}

//用于跟踪空字段的变量
设c=0；
for（设i=0；i

---

您甚至可以通过使用包含这些值的数组来简化代码

// variable for keeping track of empty field
let c = 0;
// array which contains the value
const val = [ 'FRANQUIAS', 'TELEVENDAS', 'OCC'];

for (let i = 0; i < result.length; i++) {
     // check counter is less than 3(update if there is more possibility or use c < val.length)
     // and check value is empty
     if (c < 3 && result[i].mkp == ' ') {
        // update value with corresponding value in array 
        // and increment, where `c++` returns current value and increments 
        result[i].mkp = val[c++];
    }
}

//用于跟踪空字段的变量
设c=0；
//包含值的数组
const val=['frankuias'、'TELEVENDAS'、'OCC']；
for（设i=0；i

您可以使用a作为替换值，然后使用：

const input=[{mkp:'notempty'}，{mkp:'}，{mkp:'}，{mkp:'something'}，{mkp:'}，{mkp:'something'}]；
常量替换=（函数*（）{
产生‘弗兰奎亚斯’；
产生“TELEVENDAS”；
收益率‘OCC’；
})();
const result=input.map（o=>（{…o，mkp:o.mkp===''？replacement.next（）.value:o.mkp}））；
控制台日志（结果）result

数组数据在循环之前对不起，我把变量C放在for中，所以它总是0。我换了衣服，它开始工作了！谢谢你。@Rai谢谢你，伙计。。。我没注意到在他的密码里：）@MateusFernando，你也这么说了。只是我们两人几乎同时发表了评论。没有冒犯意味着buddy.with spread语法：

input.map（o=>（{…o，mkp:o.mkp===''？replacement.next（）.value:o.mkp}））啊，非常好。答案更新，谢谢@PranavCBalan！现在看起来比以前好多了：）