如何做URL解码在Java?

在Java中,我想转换这个:

https%3A%2F%2Fmywebsite%2Fdocs%2Fenglish%2Fsite%2Fmybook.do%3Frequest_type

:

https://mywebsite/docs/english/site/mybook.do&request_type

这是我目前所拥有的:

class StringUTF
{
public static void main(String[] args)
{
try{
String url =
"https%3A%2F%2Fmywebsite%2Fdocs%2Fenglish%2Fsite%2Fmybook.do" +
"%3Frequest_type%3D%26type%3Dprivate";


System.out.println(url+"Hello World!------->" +
new String(url.getBytes("UTF-8"),"ASCII"));
}
catch(Exception E){
}
}
}

但这并不正确。这些%3A%2F格式被称为什么?我如何转换它们?

517130 次浏览

你得到的字符串是application/x-www-form-urlencoded编码。

使用URLDecoder将其转换为Java字符串。

URLDecoder.decode( url, "UTF-8" );

%3A%2F是URL编码字符。使用此java代码将它们转换回:/

String decoded = java.net.URLDecoder.decode(url, "UTF-8");

这与UTF-8或ASCII等字符编码没有任何关系。这里的字符串是URL编码。这种编码与字符编码完全不同。

试试这样做:

try {
String result = java.net.URLDecoder.decode(url, StandardCharsets.UTF_8.name());
} catch (UnsupportedEncodingException e) {
// not going to happen - value came from JDK's own StandardCharsets
}

Java 10在API中增加了对Charset的直接支持,这意味着不需要捕获UnsupportedEncodingException:

String result = java.net.URLDecoder.decode(url, StandardCharsets.UTF_8);

注意,字符编码(如UTF-8或ASCII)决定字符到原始字节的映射。有关字符编码的良好介绍,请参见这篇文章

这个问题已经被回答了之前(尽管这个问题是第一个!):

“你应该使用java.net.URI来做这件事,因为URLDecoder类做的是x-www-form-urlencoded解码,这是错误的(尽管它的名字,它是为表单数据)。”

正如URL类文档所述:

管理url的编解码方式推荐为 使用URI,并使用toURI ()和在这两个类之间进行转换 URI.toURL () . < / p > URLEncoderURLDecoder类也可以使用,但只能用于 HTML表单编码,这与编码方案不同

.定义在RFC2396标准

基本上:

String url = "https%3A%2F%2Fmywebsite%2Fdocs%2Fenglish%2Fsite%2Fmybook.do%3Frequest_type";
System.out.println(new java.net.URI(url).getPath());

会给你:

https://mywebsite/docs/english/site/mybook.do?request_type
import java.io.UnsupportedEncodingException;
import java.net.URISyntaxException;


public class URLDecoding {


String decoded = "";


public String decodeMethod(String url) throws UnsupportedEncodingException
{
decoded = java.net.URLDecoder.decode(url, "UTF-8");
return  decoded;
//"You should use java.net.URI to do this, as the URLDecoder class does x-www-form-urlencoded decoding which is wrong (despite the name, it's for form data)."
}


public String getPathMethod(String url) throws URISyntaxException
{
decoded = new java.net.URI(url).getPath();
return  decoded;
}


public static void main(String[] args) throws UnsupportedEncodingException, URISyntaxException
{
System.out.println(" Here is your Decoded url with decode method : "+ new URLDecoding().decodeMethod("https%3A%2F%2Fmywebsite%2Fdocs%2Fenglish%2Fsite%2Fmybook.do%3Frequest_type"));
System.out.println("Here is your Decoded url with getPath method : "+ new URLDecoding().getPathMethod("https%3A%2F%2Fmywebsite%2Fdocs%2Fenglish%2Fsite%2Fmybook.do%3Frequest"));


}


}

你可以明智地选择你的方法:)

我使用apache commons

String decodedUrl = new URLCodec().decode(url);

默认字符集是UTF-8

 try {
String result = URLDecoder.decode(urlString, "UTF-8");
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
public String decodeString(String URL)
{


String urlString="";
try {
urlString = URLDecoder.decode(URL,"UTF-8");
} catch (UnsupportedEncodingException e) {
// TODO Auto-generated catch block


}


return urlString;


}

使用java.net.URI类:

public String getDecodedURL(String encodedUrl) {
try {
URI uri = new URI(encodedUrl);
return uri.getScheme() + ":" + uri.getSchemeSpecificPart();
} catch (Exception e) {
return "";
}
}

请注意,异常处理可以做得更好,但这与本例没有太大关系。

如果它是整数值,我们也必须捕获NumberFormatException。

try {
Integer result = Integer.valueOf(URLDecoder.decode(urlNumber, "UTF-8"));
} catch (NumberFormatException | UnsupportedEncodingException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}

我也有这个问题,来这里寻求答案。但是我用了那个问题被批准的朋友的代码,没用。我尝试了一些不同的方法,并且成功了,所以我分享了下面的代码行,以防它有所帮助。

URLDecoder.decode(URLDecoder.decode(url, StandardCharsets.UTF_8)))