bash string comparison not working as expected - how to fix

Question

I am trying to check if a program is installed on a linux system by comparing command output with an expected output. I expect the program not to be present and so if program is abc then I should see:

bash: abc: command not found

But instead I am getting this output:

Shell for this session: bash
./checkprogs.sh: line 17: [: too many arguments
addgroup feature enabled test 1
./checkprogs.sh: line 15: tftp: command not found
./checkprogs.sh: line 17: [: too many arguments
tftp feature enabled test 2
2 tests performed

Here is the bash code:

ourshell="/bin/bash"
ourshell=${ourshell#*bin/}
echo "Shell for this session: $ourshell"

counter=1

# list of programs we want to test if available
progs='addgroup tftp'

for prog in $progs
do
    expected="$ourshell: $prog: command not found"
    #echo "$expected"
    progoutput=$($prog -h)
    #echo "prog output is: $progoutput"
    if [ $expected != $progoutput ]
    then
      echo "test failed"
    fi
    echo "$prog feature enabled test $counter"
    ((counter++))
done

((counter--))
echo "$counter tests performed"

How can I fix this?

On the system I am working on, the shell is bash and addgroup is present but tftp is not.

Please note that I am on an embedded platform and "expect" is not installed. Assume minimal linux installation.

EDIT: had to add space after [ and before ] - but even after fixing that still doesn't work as expected.

EDIT: sorry I am new to bash scripting...

If I quote the strings inside [ ] and add 2>&1 to progoutput then code looks like this:

progoutput=$($prog -h 2>&1)    # line 15

if [ "$expected" != "$progoutput" ]    # line 17

and just so anyone can try this:

progs='ls invalidxxx'     # line 9

then output is:

Shell for this session: bash
test failed
ls feature enabled test 1
test failed
invalidxxx feature enabled test 2
2 tests performed

Update the error logs you see after you add the spaces around `[` and `]` — Inian, Oct 03 '18 at 09:16
`progoutput=$($prog -h)` should be `progoutput=$($prog -h 2>&1)` — JGK, Oct 03 '18 at 09:20
Could you also quote the variables as `if [ "$expected" != "$progoutput" ]` — Inian, Oct 03 '18 at 09:21

score 1 · Accepted Answer · answered Oct 03 '18 at 09:21

1

actual method of string comparison

if [ "$expected" != "$progoutput" ]

or

if [[ $expected != $progoutput ]]

answered Oct 03 '18 at 09:21

Abhijit Pritam Dutta

5,521
2
11
17

That is a partial fix. But if you run yourself with eg line : progs='ls invalidxxx' and add 2>&1 to progoutput= line still both tests fail. – Angus Comber Oct 03 '18 at 09:30
That was helpful so accepting. Also my expected line had to be changed. – Angus Comber Oct 03 '18 at 09:43

bash string comparison not working as expected - how to fix

1 Answers1