明确一点,我并不是在寻找MIME类型。

假设我有以下输入:/path/to/file/foo.txt

我想要一种方法来分解这个输入,特别是扩展为.txt。在Java中有任何内置的方法来做到这一点吗?我希望避免编写自己的解析器。


当前回答

String path = "/Users/test/test.txt";
String extension = "";

if (path.contains("."))
     extension = path.substring(path.lastIndexOf("."));

返回. txt”

如果你只想要“txt”,将path.lastIndexOf(“.”)+ 1

其他回答

这是一种经过测试的方法

public static String getExtension(String fileName) {
    char ch;
    int len;
    if(fileName==null || 
            (len = fileName.length())==0 || 
            (ch = fileName.charAt(len-1))=='/' || ch=='\\' || //in the case of a directory
             ch=='.' ) //in the case of . or ..
        return "";
    int dotInd = fileName.lastIndexOf('.'),
        sepInd = Math.max(fileName.lastIndexOf('/'), fileName.lastIndexOf('\\'));
    if( dotInd<=sepInd )
        return "";
    else
        return fileName.substring(dotInd+1).toLowerCase();
}

测试用例:

@Test
public void testGetExtension() {
    assertEquals("", getExtension("C"));
    assertEquals("ext", getExtension("C.ext"));
    assertEquals("ext", getExtension("A/B/C.ext"));
    assertEquals("", getExtension("A/B/C.ext/"));
    assertEquals("", getExtension("A/B/C.ext/.."));
    assertEquals("bin", getExtension("A/B/C.bin"));
    assertEquals("hidden", getExtension(".hidden"));
    assertEquals("dsstore", getExtension("/user/home/.dsstore"));
    assertEquals("", getExtension(".strange."));
    assertEquals("3", getExtension("1.2.3"));
    assertEquals("exe", getExtension("C:\\Program Files (x86)\\java\\bin\\javaw.exe"));
}

JFileChooser怎么样?这并不简单,因为你需要解析它的最终输出…

JFileChooser filechooser = new JFileChooser();
File file = new File("your.txt");
System.out.println("the extension type:"+filechooser.getTypeDescription(file));

这是一个MIME类型…

好吧……我忘了你不想知道它的MIME类型。

下面链接中的有趣代码: http://download.oracle.com/javase/tutorial/uiswing/components/filechooser.html

/*
 * Get the extension of a file.
 */  
public static String getExtension(File f) {
    String ext = null;
    String s = f.getName();
    int i = s.lastIndexOf('.');

    if (i > 0 &&  i < s.length() - 1) {
        ext = s.substring(i+1).toLowerCase();
    }
    return ext;
}

相关问题: 我如何修剪一个文件扩展名从一个字符串在Java?

从文件名获取文件扩展名

/**
 * The extension separator character.
 */
private static final char EXTENSION_SEPARATOR = '.';

/**
 * The Unix separator character.
 */
private static final char UNIX_SEPARATOR = '/';

/**
 * The Windows separator character.
 */
private static final char WINDOWS_SEPARATOR = '\\';

/**
 * The system separator character.
 */
private static final char SYSTEM_SEPARATOR = File.separatorChar;

/**
 * Gets the extension of a filename.
 * <p>
 * This method returns the textual part of the filename after the last dot.
 * There must be no directory separator after the dot.
 * <pre>
 * foo.txt      --> "txt"
 * a/b/c.jpg    --> "jpg"
 * a/b.txt/c    --> ""
 * a/b/c        --> ""
 * </pre>
 * <p>
 * The output will be the same irrespective of the machine that the code is running on.
 *
 * @param filename the filename to retrieve the extension of.
 * @return the extension of the file or an empty string if none exists.
 */
public static String getExtension(String filename) {
    if (filename == null) {
        return null;
    }
    int index = indexOfExtension(filename);
    if (index == -1) {
        return "";
    } else {
        return filename.substring(index + 1);
    }
}

/**
 * Returns the index of the last extension separator character, which is a dot.
 * <p>
 * This method also checks that there is no directory separator after the last dot.
 * To do this it uses {@link #indexOfLastSeparator(String)} which will
 * handle a file in either Unix or Windows format.
 * <p>
 * The output will be the same irrespective of the machine that the code is running on.
 *
 * @param filename  the filename to find the last path separator in, null returns -1
 * @return the index of the last separator character, or -1 if there
 * is no such character
 */
public static int indexOfExtension(String filename) {
    if (filename == null) {
        return -1;
    }
    int extensionPos = filename.lastIndexOf(EXTENSION_SEPARATOR);
    int lastSeparator = indexOfLastSeparator(filename);
    return (lastSeparator > extensionPos ? -1 : extensionPos);
}

/**
 * Returns the index of the last directory separator character.
 * <p>
 * This method will handle a file in either Unix or Windows format.
 * The position of the last forward or backslash is returned.
 * <p>
 * The output will be the same irrespective of the machine that the code is running on.
 *
 * @param filename  the filename to find the last path separator in, null returns -1
 * @return the index of the last separator character, or -1 if there
 * is no such character
 */
public static int indexOfLastSeparator(String filename) {
    if (filename == null) {
        return -1;
    }
    int lastUnixPos = filename.lastIndexOf(UNIX_SEPARATOR);
    int lastWindowsPos = filename.lastIndexOf(WINDOWS_SEPARATOR);
    return Math.max(lastUnixPos, lastWindowsPos);
}

学分

复制自Apache FileNameUtils Class - http://grepcode.com/file/repo1.maven.org/maven2/commons-io/commons-io/1.3.2/org/apache/commons/io/FilenameUtils.java#FilenameUtils.getExtension%28java.lang.String%29

下面是一个正确处理.tar.gz的方法,即使是在目录名中有点的路径中:

private static final String getExtension(final String filename) {
  if (filename == null) return null;
  final String afterLastSlash = filename.substring(filename.lastIndexOf('/') + 1);
  final int afterLastBackslash = afterLastSlash.lastIndexOf('\\') + 1;
  final int dotIndex = afterLastSlash.indexOf('.', afterLastBackslash);
  return (dotIndex == -1) ? "" : afterLastSlash.substring(dotIndex + 1);
}

创建afterLastSlash是为了更快地查找afterLastBackslash,因为如果字符串中有一些斜杠,它就不必搜索整个字符串。

原始String中的char[]被重用,没有在那里添加垃圾,JVM可能会注意到afterLastSlash立即是垃圾,以便将其放在堆栈而不是堆上。

我的肮脏和可能最小的使用String.replaceAll:

.replaceAll("^.*\\.(.*)$", "$1")

请注意,第一个*是贪婪的,所以它会尽可能地抓取大多数可能的字符,然后只剩下最后一个点和文件扩展名。