为什么一个数字是有效的json? [英] Why is one number valid json?

问题描述

$.parseJSON("1")返回1.我希望这会引发错误，因为这看起来不像形式的有效JSON:

$.parseJSON("1") returns 1. I would expect this to throw an error because this does not seem like valid JSON of the form:

{
    "firstName": "John"
}

为什么1可以正确解析?反正有办法让它抛出错误.

Why does 1 parse correctly? Is there anyway to get this to throw an error instead.

推荐答案

解析数字

您可以使用 parseInt() .如果成功，它将返回一个数字，否则返回NaN(不是数字).

Parsing a number

You can better handle the parsing of numbers using parseInt(). It will return a number on success and NaN (Not a Number) otherwise.

var a = parseInt('23');
isNan(a); // false

var b = parseInt('ab');
isNan(b); // true

为什么它在jQuery中返回1

如果您查看jQuery方法的源代码，它将很快变得很清楚.

Why it returns 1 in jQuery

If you have a look at the source of the jQuery method it will become clear very quickly.

1. 它将检查是否存在对JSON.parse的本机支持.
2. 否则，它将创建一个匿名函数(具有字符串主体)，该函数仅返回JSON字符串中包含的数据并对其进行调用.

因此，如果您执行的是步骤2.，即使它不是真正的JSON，它也会简单地返回1.

So if in your case step 2. is executed it will simply return 1 even though it's not real JSON.

更新: 我很好奇原生的JSON.parse 将如何处理，它会做同样的事情(返回1).因此，无论采用哪种实施方式，您始终可以获得相同的结果.

UPDATE: I was curious how the native JSON.parse would handle it and it does the same thing (returning 1). So regardless of the implementation you always get the same result.

展出的图书馆: http://code.jquery.com/jquery-1.8.3.js

parseJSON: function( data ) {
    if ( !data || typeof data !== "string") {
        return null;
    }

    // Make sure leading/trailing whitespace is removed (IE can't handle it)
    data = jQuery.trim( data );

    // Attempt to parse using the native JSON parser first
    if ( window.JSON && window.JSON.parse ) {
        return window.JSON.parse( data );
    }

    // Make sure the incoming data is actual JSON
    // Logic borrowed from http://json.org/json2.js
    if ( rvalidchars.test( data.replace( rvalidescape, "@" )
        .replace( rvalidtokens, "]" )
        .replace( rvalidbraces, "")) ) {

        return ( new Function( "return " + data ) )(); // Just returns JSON data.

    }
    jQuery.error( "Invalid JSON: " + data );
},

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