How to match a folder name and use it in an if condition using grep in bash?

Question

for d in */ ; do
    cd $d
    NUM = $(echo ${PWD##*/} | grep -q "*abc*");
    if [[ "$NUM" -ne "0" ]]; then
        pwd
    fi
    cd ..
done

Here I'm trying to match a folder name to some substring 'abc' in the name of the folder and check if the output of the grep is not 0. But it gives me an error which reads that NUM: command not found

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An error was addressed in comments. NUM = $(echo ${PWD##*/} | grep -q "*abc*"); should be NUM=$(echo ${PWD##*/} | grep -q "*abc*"); .

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To clarify, the core problem would be to be able to match current directory name to a pattern.

Answer 1

You can probably simply the code to just

if grep -q "*abc*" <<< "${PWD##*/}" 2>/dev/null; then
   echo "$PWD"
   # Your rest of the code goes here
fi

You can use the exit code of the grep directly in a if-conditional without using a temporary variable here ( $NUM here). The condition will pass if grep was able to find a match. The here-string <<< , will pass the input to grep similar to echo with a pipeline. The part 2>/dev/null is to just suppress any errors ( stderr - file descriptor 2 ) if grep throws!

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As an additional requirement asked by OP, to negate the conditional check just do

if ! grep -q "*abc*" <<< "${PWD##*/}" 2>/dev/null; then