在Java中查找URI对应的协议
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find protocol corresponding to URI in Java
提问by M314
I have URI object in Java. I want to convert it to InputStream, but the conversion should depend on the protocol. I can make it this way if my URI is http://somepath.com/mysuperfile.xsl:
我在 Java 中有 URI 对象。我想将其转换为 InputStream，但转换应取决于协议。如果我的 URI 是http://somepath.com/mysuperfile.xsl：
return myURI.toURL().openConnection().getInputStream();
or this way if my uri is file:///somepath/mysuperfile.xsl:
或者这样，如果我的 uri 是file:///somepath/mysuperfile.xsl：
return new FileInputStream(Paths.get(myURI).toFile());
or maybe even some another way. I can try to check it manually, but do Java have some nice/proper way to check it, maybe using that new java.nio.*package?
或者甚至是另一种方式。我可以尝试手动检查它，但是 Java 是否有一些不错/正确的方法来检查它，也许使用那个新java.nio.*包？
采纳答案by Asaph
Every URI is defined as consisting of four parts, as follows:
[scheme name] : [hierarchical part] [[ ? query ]] [[ # fragment ]]
每个 URI 被定义为由四部分组成，如下所示：
[scheme name] : [hierarchical part] [[ ? query ]] [[ # fragment ]]
If the thing you want is the scheme name (which roughly translates to protocol), just use
如果您想要的是方案名称（大致翻译为协议），只需使用
switch ( myURI.getScheme() ) {
case "http":
return myURI.toURL().openConnection().getInputStream();
case "ftp":
// do something
case "file":
return new FileInputStream( Paths.get(myURI).toFile() );
}
http://docs.oracle.com/javase/6/docs/api/java/net/URI.html#getScheme%28%29
http://docs.oracle.com/javase/6/docs/api/java/net/URI.html#getScheme%28%29
or, if you just want to generate an InputStreamwithoutdifferentiating the scheme, simply use
或者，如果您只想生成一个InputStream而不区分方案，只需使用
return myURI.toURL().openStream();
or
或者
return myURI.toURL().openConnection().getInputStream();
(as you already did for HTTP protocol/scheme)
（正如您已经为 HTTP 协议/方案所做的那样）
回答by Daan Luttik
you can check the characters at the start of the string with the startsWith(String prefix) function, it is www. or http:// use the first method otherwise use the second method.
您可以使用startsWith(String prefix) 函数检查字符串开头的字符，它是www。或 http:// 使用第一种方法，否则使用第二种方法。
回答by Asaph
No need to special case file URIs. The same code works for either case. I just tested it with the following little scratch program:
无需特殊情况文件 URI。相同的代码适用于任何一种情况。我刚刚使用以下小程序对其进行了测试：
URIReadTest.java
URIReadTest.java
import java.io.*;
import java.net.*;
public class URIReadTest {
public static void main(String[] args) throws Exception {
URI uri = new URI("file:///tmp/test.txt");
InputStream in = uri.toURL().openConnection().getInputStream();
// or, more concisely:
// InputStream in = uri.toURL().openStream();
int b;
while((b = in.read()) != -1) {
System.out.print((char) b);
}
in.close();
}
}
Make a /tmp/test.txton your system and you'll see the contents of it printed out when you compile and run the above code.
/tmp/test.txt在你的系统上创建一个，当你编译和运行上面的代码时，你会看到它的内容被打印出来。
