从 Java 中的相对 URL 构建绝对 URL
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Building an absolute URL from a relative URL in Java
提问by Marty Pitt
I'm having trouble building an absolute URL from a relative URL without resorting to String hackery...
我在不诉诸字符串黑客的情况下从相对 URL 构建绝对 URL 时遇到问题......
Given
给定的
http://localhost:8080/myWebApp/someServlet
Inside the method:
方法内部：
public void handleRequest(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
}
What's the most "correct" way of building :
什么是最“正确”的构建方式：
http://localhost:8080/myWebApp/someImage.jpg
(Note, must be absolute, not relative)
（注意，必须是绝对的，不是相对的）
Currently, I'm doing it through building the string, but there MUST be a better way.
目前，我正在通过构建字符串来做到这一点，但必须有更好的方法。
I've looked at various combinations of new URI / URL, and I end up with
我查看了新 URI / URL 的各种组合，最终得到了
http://localhost:8080/someImage.jpg
Help greatly appreciated
非常感谢帮助
采纳答案by Hamza Yerlikaya
Using java.net.URL
使用 java.net.URL
URL baseUrl = new URL("http://www.google.com/someFolder/");
URL url = new URL(baseUrl, "../test.html");
回答by ZZ Coder
Looks like you already figured out the hard part, which is what host your are running on. The rest is easy,
看起来您已经弄清楚了困难的部分，即您正在运行的主机。剩下的就简单了
String url = host + request.getContextPath() + "/someImage.jpg";
Should give you what you need.
应该给你你需要的。
回答by JRL
How about:
怎么样：
String s = request.getScheme() + "://" + request.getServerName() + ":" + request.getServerPort() + request.getContextPath() + "/someImage.jpg";
回答by Forer
this code work will on linux, it can just combine the path, if you want more, constructor of URI could be helpful.
此代码将在linux上工作，它可以只组合路径，如果您想要更多，URI 的构造函数可能会有所帮助。
URL baseUrl = new URL("http://example.com/first");
URL targetUrl = new URL(baseUrl, Paths.get(baseUrl.getPath(), "second", "/third", "//fourth//", "fifth").toString());
if you path contain something need to escape, use URLEncoder.encodeto escape it at first.
如果您的路径包含需要URLEncoder.encode转义的内容，请首先使用转义它。
URL baseUrl = new URL("http://example.com/first");
URL targetUrl = new URL(baseUrl, Paths.get(baseUrl.getPath(), URLEncoder.encode(relativePath, StandardCharsets.UTF_8), URLEncoder.encode(filename, StandardCharsets.UTF_8)).toString());
example:
例子：
import java.net.MalformedURLException;
import java.net.URL;
import java.nio.file.Path;
import java.nio.file.Paths;
public class Main {
public static void main(String[] args) {
try {
URL baseUrl = new URL("http://example.com/first");
Path relativePath = Paths.get(baseUrl.getPath(), "second", "/third", "//fourth//", "fifth");
URL targetUrl = new URL(baseUrl, relativePath.toString());
System.out.println(targetUrl.toString());
} catch (MalformedURLException e) {
e.printStackTrace();
}
}
}
output
输出
http://example.com/first/second/third/fourth/fifth
baseUrl.getPath()are very important, don't forget it.
baseUrl.getPath()很重要，不要忘记。
a wrong example:
一个错误的例子：
import java.net.MalformedURLException;
import java.net.URL;
import java.nio.file.Path;
import java.nio.file.Paths;
public class Main {
public static void main(String[] args) {
try {
URL baseUrl = new URL("http://example.com/first");
Path relativePath = Paths.get("second", "/third", "//fourth//", "fifth");
URL targetUrl = new URL(baseUrl, relativePath.toString());
System.out.println(targetUrl.toString());
} catch (MalformedURLException e) {
e.printStackTrace();
}
}
}
output
输出
http://example.com/second/third/fourth/fifth
we have lost our /firstin baseurl.
我们/first在 baseurl 中丢失了。
