Question

我有一个认为1＆gt;的for循环2，走出它的循环。它应该将JSON根对象放入带有select标记的HTML option中。它必须显示两个选项，但它只在我的代码中显示一个。这是JSON：

 [{
  "batch": {
  "batch_id": 0,
"questions": [{
  "name": ["Communicatie"],
  "question_id": 25,
  "mark": ["9", "7", "10"]
    }, {
      "name": ["Samenwerking"],
  "question_id": 27,
  "mark": ["9", "8", "7"]
}, {
  "name": ["Dienstverlening"],
  "question_id": 30,
  "mark": ["9", "2", "5"]
}, {
  "name": ["Facturatie"],
  "question_id": 28,
  "mark": ["8", "7", "5"]
}, {
  "name": ["Technische kennis"],
  "question_id": 22,
  "mark": ["9", "8", "7"]
}, {
  "name": ["Distributieproces"],
  "question_id": 29,
  "mark": ["4", "7", "5"]
}, {
  "name": ["Overleg"],
  "question_id": 24,
  "mark": ["6", "8", "7"]
}, {
  "name": ["Markt kennis"],
  "question_id": 23,
  "mark": ["6", "7"]
}, {
  "name": ["Flexabiliteit"],
  "question_id": 26,
  "mark": ["3", "6", "7"]
    }]
  }
}, {
  "batch": {
"batch_id": 1,
"questions": [{
  "name": ["Communicatie"],
  "question_id": 25,
  "mark": ["7", "2"]
}, {
  "name": ["Samenwerking"],
  "question_id": 27,
  "mark": ["4", "7"]
}, {
  "name": ["Dienstverlening"],
  "question_id": 30,
  "mark": ["6", "4"]
}, {
  "name": ["Facturatie"],
  "question_id": 28,
  "mark": ["8", "3"]
}, {
  "name": ["Technische kennis"],
  "question_id": 22,
  "mark": ["6"]
}, {
  "name": ["Distributieproces"],
  "question_id": 29,
  "mark": ["3", "7"]
}, {
  "name": ["Overleg"],
  "question_id": 24,
  "mark": ["8", "4"]
}, {
  "name": ["Markt kennis"],
  "question_id": 23,
  "mark": ["3", "7"]
}, {
  "name": ["Flexabiliteit"],
  "question_id": 26,
  "mark": ["6", "2"]
}]
  }
}];

我已经隔离了一小段代码here。幸运的是，代码的这个特定部分可以工作，所以这不是问题，但我只是想让你知道我想用代码做什么。

以下是我的完整代码及评论：

 $(document).ready(function(){
    $.getJSON("/rest/BatchService/batches", function(json)
    {
        //json.length = 2, json is an arraylist of jsonobjects
        for(var i = 0; i < json.length; i++){
            //get batchid
            batchid = json[i]["batch"]["batch_id"];
            //put batchids into an option with value batchid tag and put it in select tag
            var option=$('<option />').val(json[i]["batch"]["batch_id"]).text("Batch " +json[i]["batch"]["batch_id"]);
            $("#dropdownBatches").append(option);

            //get from 'questions' the name and marks and put it in vars so that I can later put it in ChartJS
            var questions = json[i]["batch"]["questions"];
            for(var i in questions){
                var name= questions[i]["name"];
                chartjslabels.push(name);
                console.log(name);

                var marks = questions[i]["mark"];
                var sum = 0;
                for(var i = 0; i < marks.length; i++){
                    sum+= parseInt(marks[i]);
                }
                var avg = sum/marks.length;
                chartjsdata.push(avg);
            }
        }

其他一些信息：

我到处都做了断点，并且在最后一个大括号中，它假设进入下一次迭代，它停止并转到我的ChartJS来制作图表。

总而言之，我的代码中存在一个隐藏的中断，它会停止for循环。我无法找到它。

Answer 1

因为你为每个循环使用相同的变量i，所以在这里使用不同的增量[i,j,z]

$(document).ready(function(){
    $.getJSON("/rest/BatchService/batches", function(json)
    {
        //json.length = 2, json is an arraylist of jsonobjects
        for(var i = 0; i < json.length; i++){
            //get batchid
            batchid = json[i]["batch"]["batch_id"];
            //put batchids into an option tag and put it in select tag
            var option=$('<option />').val(json[i]["batch"]["batch_id"]).text("Batch " +json[i]["batch"]["batch_id"]);
            $("#dropdownBatches").append(option);

            //get from 'questions' the name and marks and put the variables in a chart
            var questions = json[i]["batch"]["questions"];
            for(var j in questions){
                var name= questions[j]["name"];
                chartjslabels.push(name);
                console.log(name);

                var marks = questions[j]["mark"];
                var sum = 0;
                for(var z = 0; z < marks.length; z++){
                    sum+= parseInt(marks[z]);
                }
                var avg = sum/marks.length;
                chartjsdata.push(avg);
            }
        }

Answer 2

您正在使用相同的变量i，将内循环更改为其他

       for(var j in questions){
            var name= questions[j]["name"];
            chartjslabels.push(name);
            console.log(name);

            var marks = questions[j]["mark"];
            var sum = 0;
            for(var k = 0; i < marks.length; i++){
                sum+= parseInt(marks[k]);
            }
            var avg = sum/marks.length;
            chartjsdata.push(avg);
        }

Answer 3

这里的问题非常简单，你在3个嵌套for循环中使用i变量，所以它的值会多次递增，导致父for循环只运行一次。

Answer 4

 for(var i in questions){

       for(var i = 0; i < marks.length; i++){

对两个循环使用i会导致您看到的行为。

Answer 5

您不能在嵌套的for循环中使用相同的变量名。这意味着i-var将在所有循环中递增，这意味着代码将不会按照您的意愿执行。

将变量名称更改为其他字母以解决问题。

我的for循环不会在Javascript中迭代

5 个答案: