CODEWITHSUNDEEP
linuxbashshell
Martynas
Martynas

Reputation: 2565

linux shell get name of file

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?

Upvotes: 2

Views: 5746

Answers (5)

Idelic
Idelic

Reputation: 15582

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

Upvotes: 0

Spencer Rathbun
Spencer Rathbun

Reputation: 14910

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.

Upvotes: 2

user unknown
user unknown

Reputation: 36259

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

Upvotes: 0

ghoti
ghoti

Reputation: 46876

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

Upvotes: 1

Karoly Horvath
Karoly Horvath

Reputation: 96306

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

Upvotes: 5

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