我需要处理传递给我们库的任何给定网址。所以一开始我是这样做的:
URL url = new URL(filename);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
url = new URL(uri.toASCIIString());
URLConnection conn = url.openConnection();
对于网址http://www.stackoverflow.com?title =Člen8FO 这给了我一个 url http://www.stackoverflow.com?title=%C4%8Clen8FO
当我发出请求时,返回了 403。于是我就这么做了:
URL url = new URL(filename);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), URLEncoder.encode(url.getQuery(), "UTF-8"), url.getRef());
url = new URL(uri.toASCIIString());
这给了我一个网址 http://www.stackoverflow.com?title%3D%C4%8Clen8FO
为什么它在显式 = 上失败?我认为这是完全正确的。
更新:这太奇怪了。下面的代码之前失败了,现在成功了。
import java.io.InputStream;
import java.net.*;
public class UriIssue {
public static void main(String[] args) {
try {
openUriWithEncode("http://www.google.com/?q=\u010clen8FO");
System.out.println("completed successfully");
} catch (Exception ex) {
System.out.println("exception = " + ex);
}
try {
openUriNoEncode("http://www.google.com/?q=\u010clen8FO");
System.out.println("completed successfully");
} catch (Exception ex) {
System.out.println("exception = " + ex);
}
}
public static void openUriWithEncode(String fullUrl) throws Exception {
System.out.println("openUriWithEncode(" + fullUrl + ")");
URL url = new URL(fullUrl);
String query = url.getQuery() != null ? URLEncoder.encode(url.getQuery(), "UTF-8") : null;
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), query, url.getRef());
System.out.println("uri = " + uri.toASCIIString());
url = new URL(uri.toASCIIString());
URLConnection conn = url.openConnection();
InputStream stream = conn.getInputStream();
stream.close();
}
public static void openUriNoEncode(String fullUrl) throws Exception {
System.out.println("openUriNoEncode(" + fullUrl + ")");
URL url = new URL(fullUrl);
URI uri = new URI(url.getProtocol(), url.getUserInfo(), url.getHost(), url.getPort(), url.getPath(), url.getQuery(), url.getRef());
System.out.println("uri = " + uri.toASCIIString());
url = new URL(uri.toASCIIString());
URLConnection conn = url.openConnection();
InputStream stream = conn.getInputStream();
stream.close();
}
}
最佳答案
您正在对包括 =
在内的整个查询进行编码,因此当然会对 =
进行编码。您只需要对键和值进行编码,例如
URLEncoder.encode(key)+"="+URLEncoder.encode(value)
关于java - 为什么需要对=进行url编码?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27284131/