在 Objective-C 中获取 NSURL 的一部分

发布于 2024-09-18 09:00:13 字数 188 浏览 7 评论 0原文

我有一个 NSString，其值为

http://digg.com/news/business/24hr

How can I get everything before the 3rd level?

http://digg.com/news/

原文

I have an NSString with the value of

http://digg.com/news/business/24hr

How can I get everything before the 3rd level?

http://digg.com/news/

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带刺的爱情 2024-09-25 09:00:13

请注意，这不完全是第三层。 URL 的分割方式如下：

协议或方案（此处为 http）、
:// 分隔符、
用户名和密码（此处没有，但它可以是username:password@hostname）
主机名（这里是digg.com）
端口（在:80之后例如域名）
路径（此处为 /news/business/24hr）
参数字符串（分号后面的任何内容）
查询字符串（如果您有 GET 参数，例如 ?foo=bar&baz=frob)
片段（如果链接中有锚点，例如 #foobar）。

一个“功能齐全”的 URL 看起来像这样：

http://foobar:[email protected]:8080/some/path/file.html;params-here?foo=bar#baz

NSURL 具有广泛的访问器。您可以在 NSURL 类，访问 URL 的各个部分部分。快速参考：

-[NSURL schema] = http
-[NSURL resourceSpecifier] =（从 // 到 URL 末尾的所有内容）
-[NSURL user ] = foobar
-[NSURL 密码] = nicate
-[NSURL 主机] = example.com
-[NSURL 端口] = 8080
第
章file.html"] （注意首字母 / 是其中的一部分）
-[NSURL lastPathComponent] = file.html
-[NSURL pathExtension] = html
- [NSURL参数字符串] = params-here
-[NSURL查询] = foo=bar
-[NSURL片段] = baz

不过，你想要什么，是这样的：

NSURL* url = [NSURL URLWithString:@"http://digg.com/news/business/24hr"];
NSString* reducedUrl = [NSString stringWithFormat:
    @"%@://%@/%@",
    url.scheme,
    url.host,
    url.pathComponents[1]];

对于您的示例 URL，您似乎想要的是协议、主机和第一个路径组件。（-[NSString pathComponents] 返回的数组中索引 0 处的元素只是“/”，因此您需要索引 1 处的元素。其他斜杠将被丢弃。）

This isn't exactly the third level, mind you. An URL is split like that way:

the protocol or scheme (here, http)
the :// delimiter
the username and the password (here there isn't any, but it could be username:password@hostname)
the host name (here, digg.com)
the port (that would be :80 after the domain name for instance)
the path (here, /news/business/24hr)
the parameter string (anything that follows a semicolon)
the query string (that would be if you had GET parameters like ?foo=bar&baz=frob)
the fragment (that would be if you had an anchor in the link, like #foobar).

A "fully-featured" URL would look like this:

http://foobar:[email protected]:8080/some/path/file.html;params-here?foo=bar#baz

NSURL has a wide range of accessors. You may check them in the documentation for the NSURL class, section Accessing the Parts of the URL. For quick reference:

-[NSURL scheme] = http
-[NSURL resourceSpecifier] = (everything from // to the end of the URL)
-[NSURL user] = foobar
-[NSURL password] = nicate
-[NSURL host] = example.com
-[NSURL port] = 8080
-[NSURL path] = /some/path/file.html
-[NSURL pathComponents] = @["/", "some", "path", "file.html"] (note that the initial / is part of it)
-[NSURL lastPathComponent] = file.html
-[NSURL pathExtension] = html
-[NSURL parameterString] = params-here
-[NSURL query] = foo=bar
-[NSURL fragment] = baz

What you'll want, though, is something like that:

NSURL* url = [NSURL URLWithString:@"http://digg.com/news/business/24hr"];
NSString* reducedUrl = [NSString stringWithFormat:
    @"%@://%@/%@",
    url.scheme,
    url.host,
    url.pathComponents[1]];

For your example URL, what you seem to want is the protocol, the host and the first path component. (The element at index 0 in the array returned by -[NSString pathComponents] is simply "/", so you'll want the element at index 1. The other slashes are discarded.)

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