Having trouble greping a variable

Shell Programming and Scripting
1

I'm trying to take a users input, and then such a text file to see if it contains it.
I get an error when running it:
"./Login.sh: Line 8: [grep: command not found
Failed to login."

echo Username
read username

if [grep $username register.txt ]; then
echo Login successful.
else
echo Failed to login.
fi

If someone could give me some input to where I'm going wrong it'd be greatly appreciated.

2

you don't need to use [ ] at all. but the problem is that you need a space after '['. if takes a command. grep is a command. you can simply use grep there.

if grep -q "$username" register.txt; then
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3 
 
echo Username
read username
grep $username register.txt > /dev/null 2>&1
if [ "$?" -eq "0" ]; then
echo Login successful.
else
echo Failed to login.
fi

$? is a speacial variable which holds the exit code of the previous command.

so, if the grep is success, then it is 0 otherwise not zero

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4

This worked, thank you.

5

Check if you have the -q flag for grep so no output is produced. That is a neater way of writing it. You may need to check you can read the file first though:-

if [ -r register.txt ]
then
   grep -q $username register.txt
   RC=$?
else
   echo "Cannot read file"
   RC=99
fi

if [ $RC -eq 0 ]
then
   echo "Login successful."
else
   echo "Failed to login."
fi

you have also tried to use the numeric test -eq to compare two strings, so results can be unpredictable on different platforms. As we are sure that 0 is numeric, don't quote it. Similarly, RC will always be numeric.

I hope that this is useful.

Robin
Liverpool/Blackburn
UK

1 Like