Java 6 中的 URL 解码
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URL Decode in Java 6
提问by danny.lesnik
I see that java.net.URLDecoder.decode(String)is deprecated in 6.
我看到它java.net.URLDecoder.decode(String)在 6 中已被弃用。
I have the following String:
我有以下字符串：
String url ="http://172.20.4.60/jsfweb/cat/%D7%9C%D7%97%D7%9E%D7%99%D7%9D_%D7%A8%D7%92%D7%99%D7%9C%D7%99%D7%9"
How should I decode it in Java 6?
我应该如何在 Java 6 中解码它？
采纳答案by Tim Cooper
Now you need to specify the character encoding of your string. Based off the information on the URLDecoderpage:
现在您需要指定字符串的字符编码。根据URLDecoder页面上的信息：
Note: The World Wide Web Consortium Recommendation states that UTF-8 should be used. Not doing so may introduce incompatibilites.
The following should work for you:
java.net.URLDecoder.decode(url, "UTF-8");
注意：万维网联盟建议指出应使用 UTF-8。不这样做可能会导致不兼容。
以下应该对您有用：
java.net.URLDecoder.decode(url, "UTF-8");
Please see Draemon's answerbelow.
请参阅下面的Draemon 的回答。
回答by Joachim Sauer
As the documentationmentions, decode(String)is deprecated because it always uses the platform default encoding, which is often wrong.
正如文档中提到的，decode(String)不推荐使用，因为它总是使用平台默认编码，这通常是错误的。
Use the two-argument version instead. You will need to specify the encoding used n the escaped parts.
改用两个参数的版本。您需要指定在转义部分中使用的编码。
回答by Mathias Schwarz
Only the decode(String)method is deprecated. You should use the decode(String, String)method to explicitly set a character encoding for decoding.
仅decode(String)弃用该方法。您应该使用该decode(String, String)方法显式设置用于解码的字符编码。
回答by Draemon
You should use java.net.URIto do this, as the URLDecoder class does x-www-form-urlencoded decoding which is wrong (despite the name, it's for form data).
您应该使用java.net.URI来执行此操作，因为 URLDecoder 类执行 x-www-form-urlencoded 解码是错误的（尽管名称如此，但它用于表单数据）。
回答by DKroot
As noted by previous posters, you should use java.net.URI class to do it:
正如之前的海报所指出的，您应该使用 java.net.URI 类来做到这一点：
System.out.println(String.format("Decoded URI: '%s'", new URI(url).getPath()));
What I want to note additionally is that if you have a path fragment of a URI and want to decode it separately, the same approach with one-argument constructor works, but if you try to use four-argument constructor it does not:
我还要注意的是，如果您有一个 URI 的路径片段并想单独对其进行解码，则使用单参数构造函数的相同方法可以工作，但是如果您尝试使用四参数构造函数，则不会：
String fileName = "Map%20of%20All%20projects.pdf";
URI uri = new URI(null, null, fileName, null);
System.out.println(String.format("Not decoded URI *WTF?!?*: '%s'", uri.getPath()));
This was tested in Oracle JDK 7. The fact that this does not work is counter-intuitive, runs contrary to JavaDocs and it should be probably considered a bug.
这在 Oracle JDK 7 中进行了测试。事实上，这不起作用是违反直觉的，与 JavaDocs 相悖，它可能应该被视为一个错误。
It could trip people who are trying to use an approach symmetrical to encoding. As noted for example in this post: "how to encode URL to avoid special characters in java", in order to encodeURI, it's a good idea to construct a URI by passing different URI parts separately since different encoding rules apply to different parts:
它可能会绊倒那些试图使用与编码对称的方法的人。正如这篇文章中提到的例子：“如何编码 URL 以避免 java 中的特殊字符”，为了对URI进行编码，最好通过分别传递不同的 URI 部分来构造 URI，因为不同的编码规则适用于不同的部分：
String fileName2 = "Map of All projects.pdf";
URI uri2 = new URI(null, null, fileName2, null);
System.out.println(String.format("Encoded URI: '%s'", uri2.toASCIIString()));
