为什么 JSON.parse 会因为空字符串而失败？

为什么:

JSON.parse('');

产生错误？

Uncaught SyntaxError: Unexpected end of input

如果它只返回 null不是更符合逻辑吗？

编辑: 这不是链接问题的副本。虽然最小有效 json 的主题与这个问题有关，但它并没有得到“为什么”。

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JSON.parse expects valid notation inside a string, whether that be object {}, array [], string "" or number types (int, float, doubles).

If there is potential for what is parsing to be an empty string then the developer should check for it.

If it was built into the function it would add extra cycles, since built in functions are expected to be extremely performant, it makes sense to not program them for the race case.

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Because '' is not a valid Javascript/JSON object. An empty object would be '{}'

For reference: https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/JSON/parse

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For a valid JSON string at least a "{}" is required. See more at the http://json.org/

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最佳答案

As an empty string is not valid JSON it would be incorrect for JSON.parse('') to return null because "null" is valid JSON. e.g. 

JSON.parse("null");

returns null. It would be a mistake for invalid JSON to also be parsed to null.

While an empty string is not valid JSON two quotes is valid JSON. This is an important distinction.

Which is to say a string that contains two quotes is not the same thing as an empty string.

JSON.parse('""');

will parse correctly, (returning an empty string). But 

JSON.parse('');

will not.

Valid minimal JSON strings are

The empty object '{}'

The empty array '[]'

The string that is empty '""'

A number e.g. '123.4'

The boolean value true 'true'

The boolean value false 'false'

The null value 'null'

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Use try-catch to avoid it:

var result = null;
try {
// if jQuery
result = $.parseJSON(JSONstring);
// if plain js
result = JSON.parse(JSONstring);
}
catch(e) {
// forget about it :)
}

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