创建 Java URL 时需要转义 '%'

Question

我在创建有效的 URL 时遇到问题，因为它需要包含一些特殊字符。

String address = "...";
URL url = new URL(address);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = uri.toURL();
InputStream input = url.openStream();
....

问题是我的地址需要包含一个&作为字符串中的一个字符，表示为%26。当我将地址创建为：

String address = "http://foo.com/?animal=monkey&banana=fruit%20%26%20delicious";

... 变成

"http://foo.com/?animal=monkey&banana=fruit%20%2526%20delicious"

在 URL-to-URI-and-back 过程中。

解决此问题的最佳方法是什么？我是否在 URL/URI 转换中使用了最佳实践？如果是这样，有没有办法在 26 之前转义“%”，以便它最后仍然是 %26？

MCVE:

import java.util.*;
import java.io.*;
import java.net.*;

public class Test {
    public static void main(String[] args) {
        try {
            String address = "http://foo.com/?animal=monkey&banana=fruit%20%26%20delicious";
            URL url = new URL(address);
            URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
            url = uri.toURL();
            System.out.println("url: " + url);
        } catch (Exception e) {
            System.out.println("Bad!");
        }
    }

}

航站楼 I/O：

$ javac Test.java
$ java Test
url: http://foo.com/?animal=monkey&banana=fruit%2520%2526%2520delicious

换句话说，我如何让它输出这个 URL:

url: http://foo.com/?animal=monkey&banana=fruit%20%26%20delicious

Answer 1

进一步挖掘后，我无法使用此方法创建 URL。我需要使用仅采用单个字符串的 URI 构造函数。

我使用这段代码来找出差异并突出问题：

import java.util.*;
import java.io.*;
import java.net.*;

public class Test {
    public static void main(String[] args) {
        ArrayList<String> addresses = new ArrayList<String>();
        addresses.add("http://foo.com/?animal=monkey&banana=fruit%20%26%20delicious");
        addresses.add("http://foo.com/?animal=monkey&banana=fruit %26 delicious");
        addresses.add("http://foo.com/?animal=monkey&banana=fruit & delicious");
        //addresses.add("http://foo.com/?animal=monkey&banana=fruit \%26 delicious"); this uses an illegal escape character

        for (String address : addresses) {
            System.out.println("~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~");
            try {
                URL url = new URL(address);
                URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
                URL urlA = uri.toURL();
                System.out.println("address: " + address);
                System.out.println("yields: ");
                System.out.println("  * urlA: " + urlA);
                uri = new URI(address);
                URL urlB = uri.toURL();
                System.out.println("  * urlB: " + urlB);
            } catch (Exception e) {
                System.out.println("  Problem!");
            }
        }
    }

}

在终端中产生了这个输出：

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
address: http://foo.com/?animal=monkey&banana=fruit%20%26%20delicious
yields: 
  * urlA: http://foo.com/?animal=monkey&banana=fruit%2520%2526%2520delicious
  * urlB: http://foo.com/?animal=monkey&banana=fruit%20%26%20delicious
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
address: http://foo.com/?animal=monkey&banana=fruit %26 delicious
yields: 
  * urlA: http://foo.com/?animal=monkey&banana=fruit%20%2526%20delicious
  Problem!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
address: http://foo.com/?animal=monkey&banana=fruit & delicious
yields: 
  * urlA: http://foo.com/?animal=monkey&banana=fruit%20&%20delicious
  Problem!

第一次试的urlB的结果是唯一有效的。

创建 Java URL 时需要转义 '%'

Need to escape a '%' when creating a Java URL

java

url

uri

character-encoding

special-characters