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Simple bash problem - checking for files and displaying the first line

Hi, I'm having a little problem with bash. I have gone through a couple of tutorials online for bash, but the only real language I've ever really used before is basic.

This looks nearly to right to me (famous last words!) but it isn't working.

This is a test of what I am trying to do - I am trying to list the contents of a directory of all text files with a certain extension (in this case *.txt). What I want to do is test if the files exist, and if they do, display the first line of each file.

in $HOME/lyrics are 2 files called 1.txt and 2.txt, and the first line of each contains a name of a song as a test.

Code:

#!/bin/bash





if [ ls $HOME/lyrics/*txt 1>/dev/null 2>&1 ]; then

        echo "Lyrics files found:"

        echo " "

        for i in $HOME/lyrics/*txt; do

                head -n 1 $i

        done

else

        echo "no text files found."

fi

All I ever get as a result is:

Code:

[me@mymachine me]$ /bin/bash lyrics.sh

No text files found.

Can anyone please tell me what I am doing wrong with the first part of the "if" for it not to give me any results?

I would use find and a test variable like

Code:

#!/bin/bash



# test variable

found=

for file in $(find $HOME/lyrics -name "*.txt"); do

    [ -z $found ] && found=1

    head -n 1 $file

done



[ -z $found ] && echo "no text files found."

Also

Code:

if ls $HOME/lyrics/*txt 1>/dev/null 2>&1; then

will works too because the ls test is not supported with [], only the
files test or variables test work with [ ]

Thanks for the tip, but like I said, i'm learning bash from scratch and this script is based on one I found on http://linuxcommand.org, but I just can't seem to get it to work.

It was taken from a larger script, but there were no variables set for it, the only difference is this IF was inside a function on it's own, but as I don't need anything else from the script, I don't need to define it as a function?

Thanks,
Morrolan

The [ ] is a test for files, variables or result of test expression, ls is not a test expression
the [] is not testing the ls return value but instead if it is a positive test expression.

see man test ([] is an alias for test) and I edited my post, see how if ls can work

Doh!

It was the [] test expression which was doing it - that's what I get for trying to be clever!

Thanks very much!!!

Morrolan