Shell script not able to parse a string under if statement

Question

I have a simple script in which I am trying to find if make some_target exists or not. In case it does not exist then print a statement and exit 1.

#!/bin/bash
set +ex
output=$(make -n some_target 2>&1 | head -1)
echo "$output"
if [ "$output" == "*No rule to make target*" ]; then
  echo "Target is not Present"
  exit 1
else
  echo "foo"  
fi

but it is throwing an error and going into else loop

make: *** No rule to make target 'some_target'.  Stop.
test.sh: 5: [: make: *** No rule to make target 'some_target'.  Stop.: unexpected operato                                                                                                  
foo

Answer 1

I was able to find an answer with below:

#!/bin/sh
set +ex 
if make -n some_target 2>&1 | grep -m 1 "No rule to make target"; then
        echo "Target is not Present"
        exit 1
else
        echo "foo"
fi

Answer 2

Pattern matching(*) is possible in bash logical operators ==, =, and != where the operand on the right is considered a pattern. So you can write your condition:

if [[ $output == *'No rule to make target'* ]]

to check if the answer contains No rule to make target.

This said, it could be more efficient to check the return code of make?

(*) This is the same basic pattern matching as done on file names. There is also a logical operator =~ that takes a regular expression.