JSON.parse 给出一个 "undefined" 对象

Question

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() 在控制台中显示 "undefined"。根据this site，我的json是有效的。它来自 php 函数给出的 json_encode。

如果有帮助的话，最终目标是遍历这个json数组。谢谢

[编辑]

我意识到我的错误实际上是一个使用文字函数的范围问题。是的，我有时有点傻。谢谢大家的帮助！

Answer 1

这不是 String，它是一个有效的 JSON，您可以在 JavaScript:

中使用

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]));
}

Answer 2

如果你有这样的return

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"}]"

那么你可以使用JSON.parse()函数

它会解码 stringify json 数据

它会 return 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 return 已经是一个 json 对象，所以不需要使用 JSON.parse();

Answer 3

你的字符串不是json对象
而是
它是一个json数组对象（检查方括号）。
所以要获得价值，你必须运行把它变成 "for" 或 "each" 或 ...
.下面每个例子
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);
            });

JSON.parse 给出一个 "undefined" 对象

JSON.parse is giving an "undefined" object

javascript

arrays

parsing

json

object-literal