谁能告诉我如何在没有扩展名的情况下获取文件名? 例子:

fileNameWithExt = "test.xml";
fileNameWithOutExt = "test";

当前回答

最简单的方法是使用正则表达式。

fileNameWithOutExt = "test.xml".replaceFirst("[.][^.]+$", "");

上面的表达式将删除最后一个点后面跟一个或多个字符。这是一个基本的单元测试。

public void testRegex() {
    assertEquals("test", "test.xml".replaceFirst("[.][^.]+$", ""));
    assertEquals("test.2", "test.2.xml".replaceFirst("[.][^.]+$", ""));
}

其他回答

请看下面的测试程序:

public class javatemp {
    static String stripExtension (String str) {
        // Handle null case specially.

        if (str == null) return null;

        // Get position of last '.'.

        int pos = str.lastIndexOf(".");

        // If there wasn't any '.' just return the string as is.

        if (pos == -1) return str;

        // Otherwise return the string, up to the dot.

        return str.substring(0, pos);
    }

    public static void main(String[] args) {
        System.out.println ("test.xml   -> " + stripExtension ("test.xml"));
        System.out.println ("test.2.xml -> " + stripExtension ("test.2.xml"));
        System.out.println ("test       -> " + stripExtension ("test"));
        System.out.println ("test.      -> " + stripExtension ("test."));
    }
}

输出:

test.xml   -> test
test.2.xml -> test.2
test       -> test
test.      -> test

如果你不喜欢导入完整的apache.commons,我提取了相同的功能:

public class StringUtils {
    public static String getBaseName(String filename) {
        return removeExtension(getName(filename));
    }

    public static int indexOfLastSeparator(String filename) {
        if(filename == null) {
            return -1;
        } else {
            int lastUnixPos = filename.lastIndexOf(47);
            int lastWindowsPos = filename.lastIndexOf(92);
            return Math.max(lastUnixPos, lastWindowsPos);
        }
    }

    public static String getName(String filename) {
        if(filename == null) {
            return null;
        } else {
            int index = indexOfLastSeparator(filename);
            return filename.substring(index + 1);
        }
    }

    public static String removeExtension(String filename) {
        if(filename == null) {
            return null;
        } else {
            int index = indexOfExtension(filename);
            return index == -1?filename:filename.substring(0, index);
        }
    }

    public static int indexOfExtension(String filename) {
        if(filename == null) {
            return -1;
        } else {
            int extensionPos = filename.lastIndexOf(46);
            int lastSeparator = indexOfLastSeparator(filename);
            return lastSeparator > extensionPos?-1:extensionPos;
        }
    }
}
public static String getFileExtension(String fileName) {
        if (TextUtils.isEmpty(fileName) || !fileName.contains(".") || fileName.endsWith(".")) return null;
        return fileName.substring(fileName.lastIndexOf(".") + 1);
    }

    public static String getBaseFileName(String fileName) {
        if (TextUtils.isEmpty(fileName) || !fileName.contains(".") || fileName.endsWith(".")) return null;
        return fileName.substring(0,fileName.lastIndexOf("."));
    }

流利的方式:

public static String fileNameWithOutExt (String fileName) {
    return Optional.of(fileName.lastIndexOf(".")).filter(i-> i >= 0)
            .filter(i-> i > fileName.lastIndexOf(File.separator))
            .map(i-> fileName.substring(0, i)).orElse(fileName);
}

简单起见,使用Java的String.replaceAll()方法,如下所示:

String fileNameWithExt = "test.xml";
String fileNameWithoutExt
   = fileNameWithExt.replaceAll( "^.*?(([^/\\\\\\.]+))\\.[^\\.]+$", "$1" );

当filenamewitheext包含完全限定路径时,这也可以工作。