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Javascript SyntaxError的jQuery parseJSON帮助_Javascript_Jquery_Json - Fatal编程技术网

Javascript SyntaxError的jQuery parseJSON帮助

Javascript SyntaxError的jQuery parseJSON帮助,javascript,jquery,json,Javascript,Jquery,Json,我有如下JSON值: {"223":[{"virtuemart_state_id":"1","state_name":"Alabama"},{"virtuemart_state_id":"2","state_name":"Alaska"}]} <script type="text/javascript" src="jquery-1.11.1.min.js"></script> <script type="text/javascript"> $("do

我有如下JSON值:

{"223":[{"virtuemart_state_id":"1","state_name":"Alabama"},{"virtuemart_state_id":"2","state_name":"Alaska"}]}
<script type="text/javascript" src="jquery-1.11.1.min.js"></script>
<script type="text/javascript">
    $("document").ready(function() {
        var state,
        url = 'http://localhost/jquery/test.json';

      $.getJSON(url, function(data){
        console.log(data);
        $.each(data.223, function(i, rep){
            state += "<option value = '" + rep.virtuemart_state_id + "'>" + rep.state_name + "</option>";

        });
        $("#state").html(state);
      });
    });
</script>
</head>
<div id="result">
    <select id="state">
    </select>
</div>
&我正在尝试解析如下数据:

{"223":[{"virtuemart_state_id":"1","state_name":"Alabama"},{"virtuemart_state_id":"2","state_name":"Alaska"}]}
<script type="text/javascript" src="jquery-1.11.1.min.js"></script>
<script type="text/javascript">
    $("document").ready(function() {
        var state,
        url = 'http://localhost/jquery/test.json';

      $.getJSON(url, function(data){
        console.log(data);
        $.each(data.223, function(i, rep){
            state += "<option value = '" + rep.virtuemart_state_id + "'>" + rep.state_name + "</option>";

        });
        $("#state").html(state);
      });
    });
</script>
</head>
<div id="result">
    <select id="state">
    </select>
</div>

$(“文档”).ready(函数(){
var州,
url='1〕http://localhost/jquery/test.json';
$.getJSON(url、函数(数据){
控制台日志(数据);
$。每个(数据223,函数(i,代表){
州+=“”+rep.state_name+“”;
});
$(“#state”).html(state);
});
});
但它不适用于数字223&我得到的错误如下:
SyntaxError:missing)在参数列表之后

知道我哪里出错了吗?谢谢

您不能使用
运算符引用名称不是有效标识符的对象属性。相反,您应该这样做:

    $.each(data[223], function(i, rep){

data.223
不是有效的Javascript。它必须是
数据['223']
快捷方式符号很方便,但它无法处理您在JS中实际创建的所有可能的键名。

尝试以下操作:
$。每个(数据[223]
。您不能引用对象属性