Question

我有以下JSON结构：

[{ "id":"10", "class": "child-of-9" }, { "id": "11", "classd": "child-of-10" }]

如何使用JavaScript迭代它？

Answer 1

var arr = [ {"id":"10", "class": "child-of-9"}, {"id":"11", "classd": "child-of-10"}];

for (var i = 0; i < arr.length; i++){
    var obj = arr[i];
    for (var key in obj){
        var attrName = key;
        var attrValue = obj[key];
    }
}

var arr = [ {"id":"10", "class": "child-of-9"}, {"id":"11", "class": "child-of-10"}];
    
for (var i = 0; i < arr.length; i++){
  document.write("<br><br>array index: " + i);
  var obj = arr[i];
  for (var key in obj){
    var value = obj[key];
    document.write("<br> - " + key + ": " + value);
  }
}

注意：for-in方法对于简单对象来说很酷。使用DOM对象不是很聪明。

Answer 2

取自jQuery docs：

var arr = [ "one", "two", "three", "four", "five" ];
var obj = { one:1, two:2, three:3, four:4, five:5 };

jQuery.each(arr, function() {
  $("#" + this).text("My id is " + this + ".");
  return (this != "four"); // will stop running to skip "five"
});

jQuery.each(obj, function(i, val) {
  $("#" + i).append(document.createTextNode(" - " + val));
});

Answer 3

使用 foreach ：

<html>
<body>
<script type="text/javascript">
var mycars = [{name:'Susita'}, {name:'BMW'}];
for (i in mycars)
{
  document.write(mycars[i].name + "<br />");
}
</script>
</body>
</html>

将导致：

Susita
BMW

Answer 4

如果不容易，请告诉我：

var jsonObject = {
  name: 'Amit Kumar',
  Age: '27'
};

for (var prop in jsonObject) {
  alert("Key:" + prop);
  alert("Value:" + jsonObject[prop]);
}

Answer 5

如果这是您的dataArray：

var dataArray = [{"id":28,"class":"Sweden"}, {"id":56,"class":"USA"}, {"id":89,"class":"England"}];

然后：

$(jQuery.parseJSON(JSON.stringify(dataArray))).each(function() {  
         var ID = this.id;
         var CLASS = this.class;
});

Answer 6

mootools示例：

var ret = JSON.decode(jsonstr);

ret.each(function(item){
    alert(item.id+'_'+item.classd);
});

Answer 7

从http://www.w3schools.com复制并粘贴，不需要JQuery开销。

var person = {fname:"John", lname:"Doe", age:25};

var text = "";
var x;
for (x in person) {
    text += person[x];
}

结果：John Doe 25

Answer 8

您可以使用像objx这样的迷你图书馆 - http://objx.googlecode.com/

您可以编写如下代码：

var data =  [ {"id":"10", "class": "child-of-9"},
              {"id":"11", "class": "child-of-10"}];

// alert all IDs
objx(data).each(function(item) { alert(item.id) });

// get all IDs into a new array
var ids = objx(data).collect("id").obj();

// group by class
var grouped = objx(data).group(function(item){ return item.class; }).obj()

有更多“插件”可供您处理此类数据，请参阅http://code.google.com/p/objx-plugins/wiki/PluginLibrary

Answer 9

对于嵌套对象，可以通过递归函数检索它：

function inside(events)
  {
    for (i in events) {
      if (typeof events[i] === 'object')
        inside(events[i]);
      else
      alert(events[i]);
    }
  }
  inside(events);

其中事件是json对象。

Answer 10

这是一个纯注释的JavaScript示例。

  <script language="JavaScript" type="text/javascript">
  function iterate_json(){
            // Create our XMLHttpRequest object
            var hr = new XMLHttpRequest();
            // Create some variables we need to send to our PHP file
            hr.open("GET", "json-note.php", true);//this is your php file containing json

            hr.setRequestHeader("Content-type", "application/json", true);
            // Access the onreadystatechange event for the XMLHttpRequest object
            hr.onreadystatechange = function() {
                if(hr.readyState == 4 && hr.status == 200) {
                    var data = JSON.parse(hr.responseText);
                    var results = document.getElementById("myDiv");//myDiv is the div id
                    for (var obj in data){
                    results.innerHTML += data[obj].id+ "is"+data[obj].class + "<br/>";
                    }
                }
            }

            hr.send(null); 
        }
</script>
<script language="JavaScript" type="text/javascript">iterate_json();</script>// call function here

Answer 11

使用jQuery时，Marquis Wang可能是最好的答案。

使用JavaScript forEach方法，纯JavaScript中的内容非常类似。 forEach将函数作为参数。然后将为数组中的每个项调用该函数，并将所述项作为参数。

简单易行：

<script>
var results = [ {"id":"10", "class": "child-of-9"}, {"id":"11", "classd": "child-of-10"}];

results.forEach( function( item ) {
    console.log( item );
    });
</script>

Answer 12

导航JSON文档的另一个解决方案是JSONiq（在Zorba引擎中实现），您可以在其中编写类似的内容：

jsoniq version "1.0";

let $doc := [
  {"id":"10", "class": "child-of-9"},
  {"id":"11", "class": "child-of-10"}
]
for $entry in members($doc)             (: binds $entry to each object in turn :)
return $entry.class                     (: gets the value associated with "class" :)

您可以在http://try.zorba.io/

上运行它

Answer 13

var jsonString = "{\"schema\": {\"title\": \"User Feedback\", \"description\":\"so\", \"type\":\"object\", \"properties\":{\"name\":{\"type\":\"string\"}}}," +
                        "\"options\":{ \"form\":{\"attributes\":{}, \"buttons\":{ \"submit\":{ \"title\":\"It\", \"click\":\"function(){alert('hello');}\" }}} }}";
var jsonData = JSON.parse(jsonString);

function Iterate(data)
{
    jQuery.each(data, function (index, value) {
        if (typeof value == 'object') {
            alert("Object " + index);
            Iterate(value);
        }
        else {
             alert(index + "   :   " + value);
        }
    });

};

Iterate(jsonData);

如何迭代JSON结构？

13 个答案: