If grep finds what it is looking for do X else Y

Question

In this statement I am trying to match if a version ($var2) exist in the path /app/$var1 (application name)

if 
find /app/$var1 -maxdepth 1 -type l -o -type d | grep $var2 #results in a nice list where i can manually get a true match.
# if match is found then execute command.
$realcmd "$@"
    rc=$?
    exit $rc
else
echo "no match, the version you are looking for does not exist"
fi

current code: this include all my code (not cleaned). command I run: "./xmodule load firefox/3.6.12" this version does exit

#!/bin/bash
# hook for some commands

#echo $@  #value of sting that is entered after "xmodule"

cmd=$(basename "$0")
#echo "called as $cmd"

if [[ $cmd = "xmodule" ]]
then
    realcmd='/app/modules/0/bin/modulecmd tcsh'

    # verify parameters
fi
# check if $@ contains value "/" to determine if a specific version is requested.
case "$@" in
*/*)
    echo "has slash"
var1=$(echo "$@" | grep -Eio '\s\w*') # Gets the aplication name and put it into var1
echo $var1   # is not printed should be "firefox"
var2=$(echo "$@" | grep -o '[^/]*$')  # Gets version name and put it into var2 
echo $var2 
# Checking if version number exist in /app/appname/
if find /app/$var1 -noleaf -maxdepth 1 -type l -o -type d | grep $var2; then
    $realcmd "$@"
    exit $?
else
    echo "no match, the version you are looking for does not exist"
    # Should there be an exit here?
fi
    ;;

*)
    echo "doesn't have a slash"
    ;;
esac

output: mycomputer [9:55am] [user/Desktop/script] -> ./xmodule load firefox/3.6.12 'has slash

3.6.12 no match, the version you are looking for does not exist

Where there is a blank (above 3.6.1) there should be the application name. I am now realizing that this must be my problem sins the path that it uses i likely just /app. But I do not think I changed anything in that part of the code.

Gordon Davisson · Accepted Answer · 2018-01-27T18:56:42.227

31

You can use the entire grep pipeline as the condition of the if statement. Use grep -q to keep it from printing the match it finds (unless you want that printed). I also simplified the exit (there's no need to store $? in a variable if you're just going to use it immediately). Here's the result:

if find "/app/$var1" -maxdepth 1 -type l -o -type d | grep -q "$var2"; then
    $realcmd "$@"
    exit $?
else
    echo "no match, the version you are looking for does not exist"
    # Should there be an exit here?
fi

BTW, since you're going to exit immediately after $realcmd, you could use exec $realcmd "$@" to replace the shell with $realcmd instead of running $realcmd as a subprocess.

edited Jan 27 '18 at 18:56

answered Oct 23 '12 at 06:54

Gordon Davisson

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When I have grep -q it goes to else, if i remove "-q" it works. – Fredrik Oct 23 '12 at 07:40
hmm now i can no longer get it to work att all. posting updated code in original question. – Fredrik Oct 23 '12 at 07:57
you could also `exec $realcmd "$@"` and omit `exit $?` entirely – glenn jackman Oct 23 '12 at 10:11

paxdiablo · Answer 2 · 2020-06-28T23:44:14.613

19

From the grep manpage:

The exit status is 0 if selected lines are found, and 1 if not found. If an error occurred the exit status is 2.

In other words, immediately following your blah blah | grep $var2, simply check the return value.

Since the exit code for a pipeline is the exit code for the last process in that pipeline, you can use something like:

find /app/$var1 -maxdepth 1 -type l -o -type d | grep $var2 ; greprc=$?
if [[ $greprc -eq 0 ]] ; then
    echo Found
else
    if [[ $greprc -eq 1 ]] ; then
        echo Not found
    else
        echo Some sort of error
    fi
fi

edited Jun 28 '20 at 23:44

answered Oct 23 '12 at 06:45

paxdiablo

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See https://stackoverflow.com/questions/36313216/why-is-testing-to-see-if-a-command-succeeded-or-not-an-anti-pattern – tripleee Jun 28 '20 at 12:08

If grep finds what it is looking for do X else Y

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