你如何在Java中浏览URL？ [英] How do you unescape URLs in Java?

问题描述

当我通过URL的InputStream读取xml，然后删除除url之外的所有内容时，我得到 http://cliveg.bu.edu/people/sganguly/player/%20Rang%20De%20Basanti%20-%20Tu%20Bin% 20Bataye.mp3 。

When I read the xml through a URL's InputStream, and then cut out everything except the url, I get "http://cliveg.bu.edu/people/sganguly/player/%20Rang%20De%20Basanti%20-%20Tu%20Bin%20Bataye.mp3".

正如你所看到的，有很多％20。

As you can see, there are a lot of "%20"s.

我想要的网址未转义。

有没有办法在Java中这样做而不使用第三方库？

Is there any way to do this in Java, without using a third-party library?

推荐答案

这不是未转义的XML，这是URL编码的文本。在我看来，你想在URL字符串上使用以下内容。

This is not unescaped XML, this is URL encoded text. Looks to me like you want to use the following on the URL strings.

URLDecoder.decode(url);

这将为您提供正确的文字。您提供的解码结果就是这样。

This will give you the correct text. The result of decoding the like you provided is this.

http://cliveg.bu.edu/people/sganguly/player/ Rang De Basanti - Tu Bin Bataye.mp3

％20是一个转义空格字符。为了得到上述内容，我使用了URLDecoder对象。

The %20 is an escaped space character. To get the above I used the URLDecoder object.

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