why 'ls' command printing the directory content multiple times - bash

Question

I have the following shell script in which I want to check the specific directory content on the remote machines and print them in a file.
file=serverList.csv
n=0
while [ $n -le 2 ]
do
while IFS=: read -r f1 f2
do
# echo line is stored in $line
if echo $f1 | grep -q "xx.xx.xxx";
then
ssh user#$f1 ls path/*war_* > path/$f1.txt < /dev/null; ls path/*zip_* >> path/$f1.txt < /dev/null;
ssh user#$f1 ls -d /apps/jetty*_* >> path/$f1.txt < /dev/null;
fi
done < "$file"
sleep 15
n=$(( n+1 ))
done
I am using this script inside a cron job for every 2 minute as following:
*/2 * * * * /path/myscript.sh
but somehow I am ending up with the following output file:
/apps/jetty/webapps_wars/test_new.war
path/ReleaseTest.static.zip_2020-08-05
path/ReleaseTest.static.zip_2020-08-05
path/ReleaseTest.static.zip_2020-08-05
path/jetty_xx.xx_2020-08-05
path/jetty_new
path/jetty_xx.xx_2020-08-05
path/jetty_new
I am not sure why am I getting the files in the list twice, sometimes 3 times. but I execute the shell directly from putty, it works fine. What do I need to change in order to correct this script?

Answer 1

Example:
~$ cd tmp
~/tmp$ mkdir test
~/tmp$ cd !$
cd test
~/tmp/test$ mkdir -p apps/jetty/webapp_wars/ && touch apps/jetty/webapp_wars/test_new.war
~/tmp/test$ mkdir path
~/tmp/test$ touch path/{ReleaseTest.static.zip_2020-08-05,jetty_xx.xx_2020-08-05,jetty_new}
~/tmp/test$ cd ..
~/tmp$ listpath=$(find test/path \( -name "*2020-08-05" -o -name "*new" \) )
~/tmp$ listapps=$(find test/apps/ -name "*war" )
~/tmp$ echo ${listpath[#]}" "${listapps[#]} | tr " " "\n" | sort > resultfile
~/tmp$
~/tmp$ cat resultfile
test/apps/jetty/webapp_wars/test_new.war
test/path/jetty_new
test/path/jetty_xx.xx_2020-08-05
test/path/ReleaseTest.static.zip_2020-08-05
~/tmp$ rm -rf test/ && unset listapps && unset listpath && rm resultfile
~/tmp$
This way you get only one result for each pattern you are looking for in your if...then...else block of code.
Just adapt the ssh ..... find commands and take care of quotes & parentheses but there is the easiest solution, this way you do not have to rewrite the script from scratch. And be careful on local / remote variables if you use them.

Answer 2

You really should not use ls but the fundamental problem is probably that three separate commands with three separate wildcards could match the same file three times.
Also, one of your commands is executed locally (you forgot to put ssh etc in front of the second one), so if the wildcard matches on your local computer, that would produce a result which doesn't reflect the situation on the remote server.
Try this refactoring.
file=serverList.csv
n=0
while [ $n -le 2 ]
do
while IFS=: read -r f1 f2
do
# echo line is stored in $line <- XXX this is not true
if echo "$f1" | grep -q "xx.xx.xxx";
then
ssh user#$f1 "printf '%s\n' path/*war_* path/*zip_* /apps/jetty*_*" | sort -u >path/"$f1".txt < /dev/null
fi
done < "$file"
sleep 15
n=$(( n+1 ))
done
The sort gets rid of any duplicates. This assumes none of your file names contain newlines; if they do, you'd need to use something which robustly handles them (try printf '%s\0' and sort -z but these are not portable).
ls would definitely also accept three different wildcards but like the link above explains, you really never want to use ls in scripts.