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社区首页 >问答首页 >无效的JSON,但在JSONLint上验证

无效的JSON,但在JSONLint上验证
EN

Stack Overflow用户
提问于 2014-03-31 01:19:39
回答 1查看 551关注 0票数 1

我有下面的JSON,它在JSONLint.com上验证,但是当我把它传递给JSON.parse()时,我得到了错误

代码语言:javascript
复制
SyntaxError: JSON.parse: unexpected character
...0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,...

这显然是最后一句“正确”:行

代码语言:javascript
复制
        var theJSON = JSON.parse({
    "data": [
        {
            "wrong": "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0",
            "correct": "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
        },
        {
            "wrong": "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0",
            "correct": "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
        },
        {
            "wrong": "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0",
            "correct": "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
        },
        {
            "wrong": "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0",
            "correct": "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
        }
    ]
});
EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2014-03-31 01:24:55

像许多人一样,您将JavaScript的字面语法与JSON混为一谈。这种情况经常发生,因为JSON使用JavaScript字面语法的子集,因此它看起来非常相似。然而,JSON始终是一个字符串。它是一种用于在langs/平台之间移植数据结构的序列化数据方案。

同样令人困惑的是,任何平台输出的一串JSON都可以被复制并粘贴到JavaScript中并使用。同样,这是因为共享语法。但是,将这样的输出粘贴到JavaScript中之后,就不再使用JSON了--他们现在正在用字面语法编写JavaScript。也就是说,除非您将其粘贴在引号之间并正确地转义结果字符串。但是这样做是没有意义的,因为为了最终得到你已经拥有的东西,它需要被解析。

JSON.parse()是一种非序列化数据的方法,它已经被序列化为JSON。它需要一个字符串,因为JSON是一个字符串。您正在传递一个对象(在字面语法中)。它不需要parsing...it已经是您想要的东西了。

用单引号包装对象文字会使代码工作,但是这样做是没有意义的,因为解析只会导致您已经拥有的东西。

如果将名为theJSON的变量替换为名为theObject的变量,并使其看起来如下,则编写代码会更好:

代码语言:javascript
复制
var theObject = {
    data: [
        {
            wrong: "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0",
            correct: "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
        },
        {
            wrong: "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0",
            correct: "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
        },
        {
            wrong: "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0",
            correct: "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
        },
        {
            wrong: "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0",
            correct: "0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0"
        }
    ]
};

任何代码想要使用的解析结果一旦完成,都应该是可以的。

票数 1
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/22752209

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