这是一个有效的URL，如果是这样，为什么NSURL无法识别其查询？ [英] Is this a valid URL, and if so why does NSURL not recognize its query?

问题描述

myscheme:?id=test

如果我通过了到 [NSURL URLWithString：] 并打印其 query 属性，我得到 nil ，但如果我在冒号（或两个或三个）后放一个斜杠，就可以了。

If I pass this to [NSURL URLWithString:] and print its query property, I get nil, but if I put a slash right after the colon (or two or three) it works fine.

这实际上是无效的URL还是它 NSURL 中的错误？

Is this actually an invalid URL or is it a bug in NSURL?

根据一些研究，它似乎是有效的URL。来自 http://url.spec.whatwg.org/ ：

According to some research, it appears to be a valid URL. From http://url.spec.whatwg.org/:

绝对URL必须是一个方案，后跟：，然后是方案
数据，还可以是？

An absolute URL must be a scheme, followed by ":", followed by scheme data, optionally followed by "?" and a query.

和（强调）

方案数据的语法取决于方案，通常在
旁边定义。对于相对方案，方案数据必须是
方案相对URL。 对于其他方案，规范或标准
必须在零个或更多 URL
个单位的约束内定义方案数据。

The syntax of scheme data depends on the scheme and is typically defined alongside it. For a relative scheme, scheme data must be a scheme-relative URL. For other schemes, specifications or standards must define scheme data within the constraints of zero or more URL units.

推荐答案

myscheme：？id = test 是有效的网址，并且 NSURL 通过从 + URLWithString返回非< nil 来识别它， code>。

myscheme:?id=test is a valid URL, and NSURL recognises it as such, by returning non-nil from +URLWithString:.

但是，根据RFC 1808，它的格式不符合 NSURL 识别？字符表示查询字符串，因此 -query 返回 nil 。

However, by RFC 1808, it doesn't conform in a manner such that NSURL recognises the ? character as indicating a query string, so -query returns nil.

这里的好消息是 NSURLComponents 的解析器与将识别为？表示对您的示例的查询：

The good news here is that NSURLComponents has a slightly different parser which will recognise the ? as indicating a query for your example:

NSURLComponents *components = [NSURLComponents componentsWithString:@"myscheme:?id=test"];
return components.query; // returns @"id=test"

如果您仍然需要以此为目标的iOS 6或更早版本代码，建议您使用 KSURLComponents 类而是采用相同的解析方法。

If you still need to target iOS 6 or earlier with this code, I suggest using my KSURLComponents class instead, which takes the same parsing approach.

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