JSON.parse正在提供一个" undefined"宾语

时间:2015-02-26 04:59:43

标签: javascript arrays json parsing object-literal

我试图解析这个字符串:

[{"ZoneId": "1", "0": "1", "ZoneX": "29", "1": "29", "ZoneY": "27", "2":     "27", "ZoneWidth": "76", "3": "76", "ZoneHeight": "61", "4": "61", "ZoneImage": "46", "5": "46", "ZonePointTo": "2", "6": "2"},
{"ZoneId": "2", "0": "2", "ZoneX": "382", "1": "382", "ZoneY": "226", "2": "226", "ZoneWidth": "-117", "3": "-117", "ZoneHeight": "98", "4": "98", "ZoneImage": "46", "5": "46", "ZonePointTo": "3", "6": "3"},
{"ZoneId": "3", "0": "3", "ZoneX": "108", "1": "108", "ZoneY": "74", "2": "74", "ZoneWidth": "363", "3": "363", "ZoneHeight": "83", "4": "83", "ZoneImage": "46", "5": "46", "ZonePointTo": "2", "6": "2"}]

在此字符串上使用JSON.parse()会在控制台中显示“undefined”。根据{{​​3}},我的json是有效的。它来自php函数给出的json_encode。

如果它可以提供帮助,最终目标是遍历这个json数组。感谢。

[编辑]

我意识到我的错误实际上是使用文字函数的范围问题。是的,我有时候有点愚蠢。谢谢大家的帮助!

3 个答案:

答案 0 :(得分:5)

这不是String,它是一个可以在JavaScript中使用的有效JSON:

var jsonData = [{"ZoneId": "1", "0": "1", "ZoneX": "29", "1": "29", "ZoneY": "27", "2":     "27", "ZoneWidth": "76", "3": "76", "ZoneHeight": "61", "4": "61", "ZoneImage": "46", "5": "46", "ZonePointTo": "2", "6": "2"},
{"ZoneId": "2", "0": "2", "ZoneX": "382", "1": "382", "ZoneY": "226", "2": "226", "ZoneWidth": "-117", "3": "-117", "ZoneHeight": "98", "4": "98", "ZoneImage": "46", "5": "46", "ZonePointTo": "3", "6": "3"},
{"ZoneId": "3", "0": "3", "ZoneX": "108", "1": "108", "ZoneY": "74", "2": "74", "ZoneWidth": "363", "3": "363", "ZoneHeight": "83", "4": "83", "ZoneImage": "46", "5": "46", "ZonePointTo": "2", "6": "2"}];

for(index in jsonData) {
    alert(JSON.stringify(jsonData[index]));
}

答案 1 :(得分:1)

您的字符串不是json对象
,但
它是json数组对象(请检查方括号)。
因此,要获取该值,必须将其输入“ for”或“ each”或...
。下面的每个示例
将字符串传递给变量,然后;

var obj=[{"ZoneId": "1", "0": "1", "ZoneX": "29", "1": "29", "ZoneY": "27", "2":     "27", "ZoneWidth": "76", "3": "76", "ZoneHeight": "61", "4": "61", "ZoneImage": "46", "5": "46", "ZonePointTo": "2", "6": "2"},
{"ZoneId": "2", "0": "2", "ZoneX": "382", "1": "382", "ZoneY": "226", "2": "226", "ZoneWidth": "-117", "3": "-117", "ZoneHeight": "98", "4": "98", "ZoneImage": "46", "5": "46", "ZonePointTo": "3", "6": "3"},
{"ZoneId": "3", "0": "3", "ZoneX": "108", "1": "108", "ZoneY": "74", "2": "74", "ZoneWidth": "363", "3": "363", "ZoneHeight": "83", "4": "83", "ZoneImage": "46", "5": "46", "ZonePointTo": "2", "6": "2"}]

            jQuery.each(obj, function(key,value) {
                alert(value.ZoneId);
            });

答案 2 :(得分:0)

如果你有这样的回报

var json_string =  "[{"0":"1","1":"29","2":"27","3":"76","4":"61","5":"46","6":"2","ZoneId":"1","ZoneX":"29","ZoneY":"27","ZoneWidth":"76","ZoneHeight":"61","ZoneImage":"46","ZonePointTo":"2"},{"0":"2","1":"382","2":"226","3":"-117","4":"98","5":"46","6":"3","ZoneId":"2","ZoneX":"382","ZoneY":"226","ZoneWidth":"-117","ZoneHeight":"98","ZoneImage":"46","ZonePointTo":"3"},{"0":"3","1":"108","2":"74","3":"363","4":"83","5":"46","6":"2","ZoneId":"3","ZoneX":"108","ZoneY":"74","ZoneWidth":"363","ZoneHeight":"83","ZoneImage":"46","ZonePointTo":"2"}]"

然后你可以使用JSON.parse()函数

它将解码stringify json数据

它会回复你

 [{"ZoneId": "1", "0": "1", "ZoneX": "29", "1": "29", "ZoneY": "27", "2":     "27", "ZoneWidth": "76", "3": "76", "ZoneHeight": "61", "4": "61", "ZoneImage": "46", "5": "46", "ZonePointTo": "2", "6": "2"},
{"ZoneId": "2", "0": "2", "ZoneX": "382", "1": "382", "ZoneY": "226", "2": "226", "ZoneWidth": "-117", "3": "-117", "ZoneHeight": "98", "4": "98", "ZoneImage": "46", "5": "46", "ZonePointTo": "3", "6": "3"},
{"ZoneId": "3", "0": "3", "ZoneX": "108", "1": "108", "ZoneY": "74", "2": "74", "ZoneWidth": "363", "3": "363", "ZoneHeight": "83", "4": "83", "ZoneImage": "46", "5": "46", "ZonePointTo": "2", "6": "2"}]

因为返回已经是一个json对象,所以不需要使用JSON.parse();