groovy 是否有一种简单的方法来获取没有扩展名的文件名？

Question

Say I have something like this:

new File("test").eachFile() { file->  
println file.getName()  
}

This prints the full filename of every file in the test directory. Is there a Groovy way to get the filename without any extension? (Or am I back in regex land?)

Answer 1

I believe the grooviest way would be:

file.name.lastIndexOf('.').with {it != -1 ? file.name[0..<it] : file.name}

or with a simple regexp:

file.name.replaceFirst(~/\.[^\.]+$/, '')

also there's an apache commons-io java lib for that kinda purposes, which you could easily depend on if you use maven:

org.apache.commons.io.FilenameUtils.getBaseName(file.name)

Answer 2

最干净的方式。

String fileWithoutExt = file.name.take(file.name.lastIndexOf('.'))

Answer 3

Simplest way is:

'file.name.with.dots.tgz' - ~/\.\w+$/​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​​

Result is:

file.name.with.dots

Answer 4

new File("test").eachFile() { file->  
    println file.getName().split("\\.")[0]
}

This works well for file names like: foo, foo.bar

But if you have a file foo.bar.jar, then the above code prints out: foo If you want it to print out foo.bar instead, then the following code achieves that.

new File("test").eachFile() { file->  
    def names = (file.name.split("\\.")
    def name = names.size() > 1 ? (names - names[-1]).join('.') : names[0]
    println name
}

Answer 5

The FilenameUtils class, which is part of the apache commons io package, has a robust solution. Example usage:

import org.apache.commons.io.FilenameUtils

String filename = '/tmp/hello-world.txt'
def fileWithoutExt = FilenameUtils.removeExtension(filename)

This isn't the groovy way, but might be helpful if you need to support lots of edge cases.

Answer 6

Maybe not as easy as you expected but working:

new File("test").eachFile { 
  println it.name.lastIndexOf('.') >= 0 ? 
     it.name[0 .. it.name.lastIndexOf('.')-1] : 
     it.name 
  }

Answer 7

As mentioned in comments, where a filename ends & an extension begins depends on the situation. In my situation, I needed to get the basename (file without path, and without extension) of the following types of files: { foo.zip , bar/foo.tgz , foo.tar.gz } => all need to produce " foo " as the filename sans extension. (Most solutions, given foo.tar.gz would produce foo.tar .)

Here's one (obvious) solution that will give you everything up to the first "."; optionally, you can get the entire extension either in pieces or (in this case) as a single remainder (splitting the filename into 2 parts). (Note: although unrelated to the task at hand, I'm also removing the path as well, by calling file.name .)

file=new File("temp/foo.tar.gz")
file.name.split("\\.", 2)[0]    // => return "foo" at [0], and "tar.gz" at [1]

Answer 8

You can use regular expressions better. A function like the following would do the trick:

def getExtensionFromFilename(filename) {
  def returned_value = ""
  m = (filename =~ /(\.[^\.]*)$/)
  if (m.size()>0) returned_value = ((m[0][0].size()>0) ? m[0][0].substring(1).trim().toLowerCase() : "");
  return returned_value
}

Answer 9

Note

import java.io.File;

def fileNames    = [ "/a/b.c/first.txt", 
                     "/b/c/second",
                     "c:\\a\\b.c\\third...",
                     "c:\\a\b\\c\\.text"
                   ]

def fileSeparator = "";

fileNames.each { 
    // You can keep the below code outside of this loop. Since my example
    // contains both windows and unix file structure, I am doing this inside the loop.
    fileSeparator= "\\" + File.separator;
    if (!it.contains(File.separator)) {
        fileSeparator    =  "\\/"
    }

    println "File extension is : ${it.find(/((?<=\.)[^\.${fileSeparator}]+)$/)}"
    it    =  it.replaceAll(/(\.([^\.${fileSeparator}]+)?)$/,"")

    println "Filename is ${it}" 
}

Some of the below solutions (except the one using apache library) doesn't work for this example - c:/test.me/firstfile

If I try to find an extension for above entry, I will get ".me/firstfile" - :(

Better approach will be to find the last occurrence of File.separator if present and then look for filename or extension.

Note: (There is a little trick happens below. For Windows, the file separator is \\. But this is a special character in regular expression and so when we use a variable containing the File.separator in the regular expression, I have to escape it. That is why I do this:

def fileSeparator= "\\" + File.separator;

Hope it makes sense :)

Try this out:

import java.io.File;

String strFilename     =  "C:\\first.1\\second.txt";
// Few other flavors 
// strFilename = "/dd/dddd/2.dd/dio/dkljlds.dd"

def fileSeparator= "\\" + File.separator;
if (!strFilename.contains(File.separator)) {
    fileSeparator    =  "\\/"
}

def fileExtension = "";
(strFilename    =~ /((?<=\.)[^\.${fileSeparator}]+)$/).each { match,  extension -> fileExtension = extension }
println "Extension is:$fileExtension"

Answer 10

// Create an instance of a file (note the path is several levels deep)
File file = new File('/tmp/whatever/certificate.crt')

// To get the single fileName without the path (but with EXTENSION! so not answering the question of the author. Sorry for that...)
String fileName = file.parentFile.toURI().relativize(file.toURI()).getPath()