linux shell get name of file

Question

I am writting shell script.

I have following files:

2012-03-08_16-37-41
2012-03-08_16-37-43
2012-03-08_16-37-46
2012-03-08_16-37-55

Simple script:

#!/bin/bash
FILENAME= ????
echo $FILENAME

And FILENAME value should be 2012-03-08_16-37-55 (last element of sorted file name list). Also, begin of file name should be 2012 .

How could I solve this problem?

Answer 1

FILENAME=$(ls -r 2012* | head -n 1)

Answer 2

Don't parse ls output. Instead, use find:

#!/bin/sh

find . -name 2012* | sort | tail -1

To assign the result to a variable:

#!/bin/sh

filename=$(find . -name 2012* | sort | tail -1)

This also gives you access to all of the many options find has , including (not)following symlinks, recursion, only returning files (not directories), checking timestamps and so on.

Answer 3

You can use either the ls command to get files, or just use "file globbing" to expand wildcards.

#!/bin/sh

for filename in 2012*; do
  echo "File (by globbing) is $filename"
done

ls 2012* | while read filename; do
  echo "File (via ls) is $filename"
done

To get the last one, the easiest way may be to sort the ls output:

filename=`ls -r 2012* | head -1`

But you can also just tail the glob, if you want to be messy.

for filename in 2012*; do
  echo "File (by globbing) is $filename"
done | tail -1

Answer 4

f=""; 
for f in 2012* ; 
do
  # haha - don't do anything. 
  dummy=42
done; 
echo "now do something with $f"

Answer 5

Without using any external commands: set 2012*; eval FILE=\\$$# set 2012*; eval FILE=\\$$# . This is a perfectly safe use of eval .