Get the last file name in a variable in bash?

Question

In a directory, I have the following files:

myfile_v01.txt
myfile_v02.txt
myfile_v03.txt

How to make a bash script that detects the value of the last version and put the filename of this last version into a variable?

Answer 1

Not very refined but seems functional:

var=$(ls | sort -V | tail -n 1)

Then you have the last file stored in $var

Answer 2

There are two obvious approaches, you can either parse the file name or look for the latest file:

1. Parse the file name, this assumes all files are named as in your example:

   $ latest=myfile_v$(
       for f in myfile_v*; do 
         ver=${f##myfile_v}; ver=${ver%.txt}; echo $ver; 
       done | sort -n | tail -n 1).txt
   

2. Get the newest file (assuming relatively sane file names that don't contain newlines)

   $ latest=$(ls -tr | tail -n 1)
   

Answer 3

Here is a solution which uses globbing (so does not rely on parsing ls), like terdon's first solution, but doesn't need to iterate over all files:

myfiles=(myfile_v*)
lastfile="${myfiles[${#myfiles[@]}-1]}"
unset myfiles

First, all matching files are read into an array. Then, the last element of the array gets assigned to $lastfile. Finally, the array isn't needed anymore, so it can be deleted.

Unfortunately, bash (at least to my knowledge) doesn't support sorting the result from globbing^#, so this example works fine with you given naming scheme (exactly two digits, also for versions < 10), but will fail for myfile_v1.txt, myfile_v2.txt, ..., myfile_v10.txt.

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^# zsh (of course ;)) does, so zsh% myfiles=(myfile_v*(n)); lastfile="${myfiles[-1]}"; unset myfiles wouldn't suffer from this limitation.