Question

我需要将所有json对象和帧循环到html中。我可以迭代所有json对象，但我不能只获得一个json对象（第一个json对象）。

var res = '[{"ID":"246","mobile":"samsung","feedback":"feedback goes here"},{"ID":"1485","mobile":"Moto","feedback":"feedback goes here"},{"ID":"6982","mobile":"iPhone","feedback":"feedback goes here"}]';
obj = JSON.parse(res);
console.log('response length:' + obj.length);
for (var i = 0; i < obj.length; i++) {
    var finalResult = "";
    var objects = obj[i];
    for (var key in objects) {
        var res = "<tr><td>" + objects.ID + "</td><td>" + objects.mobile + "</td><td>" + objects.feedback + "</td><td></tr>";
        console.log('res:' + res);
        finalResult = res.concat(res);
        console.log('finalResult:' + finalResult);
    }
}

我无法将所有内容都放入'tr'元素，因为javascript doest没有stringbuffer。我认为可以在java中使用StringBuffer。如何使用javascript / jquery完成？

请帮助我。

Answer 1

您正在代码中间重新定义obj：

 var obj = obj[i];

使用其他名称：

你也在循环中重置你的finalResult（并且不需要内部循环）：

http://jsfiddle.net/7j52myca/

var res = '[{"ID":"246","mobile":"samsung","feedback":"feedback goes here"},{"ID":"1485","mobile":"Moto","feedback":"feedback goes here"},{"ID":"6982","mobile":"iPhone","feedback":"feedback goes here"}]';
var phones = JSON.parse(res);
console.log('response length:' + phones.length);
var finalResult = "";
for (var i = 0; i < phones.length; i++) {
    var str1 = "aasd";
    var obj = phones[i];
    var res = "<tr><td>" + obj.ID + "</td><td>" + obj.mobile + "</td><td>" + obj.feedback + "</td><td></tr>";
    console.log('res:' + res);
    finalResult += res;
}
console.log('finalResult:' + finalResult);
$('#result').append(finalResult);

你可以使用纯粹的jQuery做同样的事情，但是@Rory McCrossan已经发布了一个很好的版本，所以我不打算在这里添加一个。

Answer 2

问题是因为您正在重新定义您在迭代函数中循环的变量obj。

正如您使用jQuery对此进行了标记，因此使用$.each来构建表格是一个较短的替代方法：

＆＃13;

var res = '[{"ID":"246","mobile":"samsung","feedback":"feedback goes here"},{"ID":"1485","mobile":"Moto","feedback":"feedback goes here"},{"ID":"6982","mobile":"iPhone","feedback":"feedback goes here"}]';
var phones = JSON.parse(res);

$.each(phones, function(i, obj) {
    $('<tr />')
        .append('<td>' + obj.ID + '</td><td>' + obj.mobile + '</td><td>' + obj.feedback + '</td><td>')
        .appendTo('table');
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table></table>

＆＃13;

Answer 3

您不需要内部for循环，您可以直接访问您的元素。我还做了一些其他的改动，可以在下面找到：

var res = '[{"ID":"246","mobile":"samsung","feedback":"feedback goes here"},{"ID":"1485","mobile":"Moto","feedback":"feedback goes here"},{"ID":"6982","mobile":"iPhone","feedback":"feedback goes here"}]';

obj = JSON.parse(res);
var finalResult = "";

console.log('response length:' + obj.length);

for (var i = 0; i < obj.length; i++) {

    var str1 = "aasd";
    
    var tableRow = "<tr><td>" + obj[i]["ID"] + "</td><td>" + obj[i]["mobile"] + "</td><td>" + obj[i]["feedback"] + "</td><td></tr>";

    finalResult += tableRow;

}

console.log('finalResult:' + finalResult);

我希望你觉得它很有帮助。

Answer 4

使用jquery你可以做的每种方法

var res = '[{"ID":"246","mobile":"samsung","feedback":"feedback goes here"},{"ID":"1485","mobile":"Moto","feedback":"feedback goes here"},{"ID":"6982","mobile":"iPhone","feedback":"feedback goes here"}]';
obj = JSON.parse(res);

var finalResult = "";
$.each(obj, function(i, item) {

  var res = "<tr><td>" + item.ID + "</td><td>" + item.mobile + "</td><td>" + item.feedback + "</td><td></tr>";
  console.log('res:' + res);
  finalResult += res;
});
$('body').append(finalResult)

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 5

请参阅此代码

var jsonobj = '[{"ID":"246","mobile":"samsung","feedback":"feedback goes here"},{"ID":"1485","mobile":"Moto","feedback":"feedback goes here"},{"ID":"6982","mobile":"iPhone","feedback":"feedback goes here"}]';
obj = JSON.parse(jsonobj);
alert(obj);
$.each(obj, function (index, item) {

    $('<tr/>')
        .append('<td>' + item.ID + '</td><td>' + item.mobile + '</td><td>' + item.feedback + '</td><td>')
        .appendTo('table');

});

小提琴：http://jsfiddle.net/21wrqrfb/

Answer 6

我想你想这样做：

var res = '[{"ID":"246","mobile":"samsung","feedback":"feedback goes here"},{"ID":"1485","mobile":"Moto","feedback":"feedback goes here"},{"ID":"6982","mobile":"iPhone","feedback":"feedback goes here"}]';
obj = JSON.parse(res);
console.log('response length:' + obj.length);
var finalResult = "";
for (var i = 0; i < obj.length; i++) {  
    var str1 = "aasd";
    var obj1 = obj[i];
    for (var key in obj1) {
        var res = "<tr><td>" + obj1.ID + "</td><td>" + obj1.mobile + "</td><td>" + obj1.feedback + "</td><td></tr>";
        //console.log('res:' + res);
        finalResult += res;
    }
}
console.log('finalResult:' + finalResult);

迭代所有JSON对象并放入html

6 个答案: