如何获取Java中不带扩展名的文件名？

Question

Can anyone tell me how to get the filename without the extension? Example:

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

Answer 1

If you, like me, would rather use some library code where they probably have thought of all special cases, such as what happens if you pass in null or dots in the path but not in the filename, you can use the following:

import org.apache.commons.io.FilenameUtils;
String fileNameWithOutExt = FilenameUtils.removeExtension(fileNameWithExt);

Answer 2

The easiest way is to use a regular expression.

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

The above expression will remove the last dot followed by one or more characters. Here's a basic unit test.

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

Answer 3

Here is the consolidated list order by my preference.

Using apache commons

import org.apache.commons.io.FilenameUtils;

String fileNameWithoutExt = FilenameUtils.getBaseName(fileName);
                          
                           OR

String fileNameWithOutExt = FilenameUtils.removeExtension(fileName);

Using Google Guava (If u already using it)

import com.google.common.io.Files;
String fileNameWithOutExt = Files.getNameWithoutExtension(fileName);

Files.getNameWithoutExtension

Or using Core Java

1)

String fileName = file.getName();
int pos = fileName.lastIndexOf(".");
if (pos > 0 && pos < (fileName.length() - 1)) { // If '.' is not the first or last character.
    fileName = fileName.substring(0, pos);
}

if (fileName.indexOf(".") > 0) {
   return fileName.substring(0, fileName.lastIndexOf("."));
} else {
   return fileName;
}

private static final Pattern ext = Pattern.compile("(?<=.)\\.[^.]+$");

public static String getFileNameWithoutExtension(File file) {
    return ext.matcher(file.getName()).replaceAll("");
}

Liferay API

import com.liferay.portal.kernel.util.FileUtil; 
String fileName = FileUtil.stripExtension(file.getName());

Answer 4

See the following test program:

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."));
    }
}

which outputs:

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

Answer 5

If your project uses Guava (14.0 or newer), you can go with Files.getNameWithoutExtension() .

(Essentially the same as FilenameUtils.removeExtension() from Apache Commons IO, as the highest-voted answer suggests. Just wanted to point out Guava does this too. Personally I didn't want to add dependency to Commons—which I feel is a bit of a relic—just because of this.)

Answer 6

Below is reference from https://android.googlesource.com/platform/tools/tradefederation/+/master/src/com/android/tradefed/util/FileUtil.java

/**
 * Gets the base name, without extension, of given file name.
 * <p/>
 * e.g. getBaseName("file.txt") will return "file"
 *
 * @param fileName
 * @return the base name
 */
public static String getBaseName(String fileName) {
    int index = fileName.lastIndexOf('.');
    if (index == -1) {
        return fileName;
    } else {
        return fileName.substring(0, index);
    }
}

Answer 7

If you don't like to import the full apache.commons, I've extracted the same functionality:

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;
        }
    }
}

Answer 8

While I am a big believer in reusing libraries, the org.apache.commons.io JAR is 174KB, which is noticably large for a mobile app.

If you download the source code and take a look at their FilenameUtils class, you can see there are a lot of extra utilities, and it does cope with Windows and Unix paths, which is all lovely.

However, if you just want a couple of static utility methods for use with Unix style paths (with a "/" separator), you may find the code below useful.

The removeExtension method preserves the rest of the path along with the filename. There is also a similar getExtension .

/**
 * Remove the file extension from a filename, that may include a path.
 * 
 * e.g. /path/to/myfile.jpg -> /path/to/myfile 
 */
public static String removeExtension(String filename) {
    if (filename == null) {
        return null;
    }

    int index = indexOfExtension(filename);

    if (index == -1) {
        return filename;
    } else {
        return filename.substring(0, index);
    }
}

/**
 * Return the file extension from a filename, including the "."
 * 
 * e.g. /path/to/myfile.jpg -> .jpg
 */
public static String getExtension(String filename) {
    if (filename == null) {
        return null;
    }

    int index = indexOfExtension(filename);

    if (index == -1) {
        return filename;
    } else {
        return filename.substring(index);
    }
}

private static final char EXTENSION_SEPARATOR = '.';
private static final char DIRECTORY_SEPARATOR = '/';

public static int indexOfExtension(String filename) {

    if (filename == null) {
        return -1;
    }

    // Check that no directory separator appears after the 
    // EXTENSION_SEPARATOR
    int extensionPos = filename.lastIndexOf(EXTENSION_SEPARATOR);

    int lastDirSeparator = filename.lastIndexOf(DIRECTORY_SEPARATOR);

    if (lastDirSeparator > extensionPos) {
        LogIt.w(FileSystemUtil.class, "A directory separator appears after the file extension, assuming there is no file extension");
        return -1;
    }

    return extensionPos;
}

Answer 9

对于 Kotlin，它现在很简单：

val fileNameStr = file.nameWithoutExtension

Answer 10

You can use java split function to split the filename from the extension, if you are sure there is only one dot in the filename which for extension.

File filename = new File('test.txt'); File.getName().split("[.]");

so the split[0] will return "test" and split[1] will return "txt"

Answer 11

fileEntry.getName().substring(0, fileEntry.getName().lastIndexOf("."));

Answer 12

Given the String filename , you can do:

String filename = "test.xml";
filename.substring(0, filename.lastIndexOf("."));   // Output: test
filename.split("\\.")[0];   // Output: test

Answer 13

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("."));
    }

Answer 14

The fluent way:

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);
}

Answer 15

从相对路径或完整路径获取名称的最简单方法是使用

import org.apache.commons.io.FilenameUtils; FilenameUtils.getBaseName(definitionFilePath)

Answer 16

You can split it by "." and on index 0 is file name and on 1 is extension, but I would incline for the best solution with FileNameUtils from apache.commons-io like it was mentioned in the first article. It does not have to be removed, but sufficent is:

String fileName = FilenameUtils.getBaseName("test.xml");

Answer 17

Use FilenameUtils.removeExtension from Apache Commons IO

Example:

You can provide full path name or only the file name .

String myString1 = FilenameUtils.removeExtension("helloworld.exe"); // returns "helloworld"
String myString2 = FilenameUtils.removeExtension("/home/abc/yey.xls"); // returns "yey"

Hope this helps ..

Answer 18

Keeping it simple, use Java's String.replaceAll() method as follows:

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

This also works when fileNameWithExt includes the fully qualified path.

Answer 19

My solution needs the following import.

import java.io.File;

The following method should return the desired output string:

private static String getFilenameWithoutExtension(File file) throws IOException {
    String filename = file.getCanonicalPath();
    String filenameWithoutExtension;
    if (filename.contains("."))
        filenameWithoutExtension = filename.substring(filename.lastIndexOf(System.getProperty("file.separator"))+1, filename.lastIndexOf('.'));
    else
        filenameWithoutExtension = filename.substring(filename.lastIndexOf(System.getProperty("file.separator"))+1);

    return filenameWithoutExtension;
}

Answer 20

com.google.common.io.Files

Files.getNameWithoutExtension(sourceFile.getName())

can do a job as well

Answer 21

file name only, where full path is also included. No need for external libs, regex...etc

    public class MyClass {
    public static void main(String args[]) {
  
  
      String file  = "some/long/directory/blah.x.y.z.m.xml";

      System.out.println(file.substring(file.lastIndexOf("/") + 1, file.lastIndexOf(".")));
     //outputs blah.x.y.z.m
    }

}

Answer 22

Try the code below. Using core Java basic functions. It takes care of String s with extension, and without extension (without the '.' character). The case of multiple '.'is also covered.

String str = "filename.xml";
if (!str.contains(".")) 
    System.out.println("File Name=" + str); 
else {
    str = str.substring(0, str.lastIndexOf("."));
    // Because extension is always after the last '.'
    System.out.println("File Name=" + str);
}

You can adapt it to work with null strings.