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

private String getFileExtension(File file) {
    String name = file.getName();
    int lastIndexOf = name.lastIndexOf(".");
    if (lastIndexOf == -1) {
        return ""; // empty extension
    }
    return name.substring(lastIndexOf);
}

其他回答

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

在这里我做了一个小方法(然而不是那么安全,并没有检查很多错误),但如果只有你在编写一个普通的java程序,这就足够找到文件类型了。这对于复杂的文件类型并不适用,但这些文件类型通常不常用。

    public static String getFileType(String path){
       String fileType = null;
       fileType = path.substring(path.indexOf('.',path.lastIndexOf('/'))+1).toUpperCase();
       return fileType;
}

只是一个基于正则表达式的替代方案。没那么快,也没那么好。

Pattern pattern = Pattern.compile("\\.([^.]*)$");
Matcher matcher = pattern.matcher(fileName);

if (matcher.find()) {
    String ext = matcher.group(1);
}

Java 20 EA

从Java 20 EA(早期访问)开始,终于有了一个新方法Path#getExtension,它将扩展名作为字符串返回:

Paths.get("/Users/admin/notes.txt").getExtension();              // "txt"
Paths.get("/Users/admin/.gitconfig").getExtension();             // "gitconfig"
Paths.get("/Users/admin/configuration.xml.zip").getExtension();  // "zip"
Paths.get("/Users/admin/file").getExtension();                   // null

这个特别的问题给了我很多麻烦,然后我找到了一个非常简单的解决方案,我张贴在这里。

file.getName().toLowerCase().endsWith(".txt");

就是这样。