[SOLVED] Bash command - using IF - FI within a DO - DONE

Issue

I'm trying to run a command that should find PHP files that contain "base64_decode" and/or "eval", echo the file name, print the top three lines, if the file contains more than 3 lines, also the bottom 3.

I have the following at the moment:

for file in $(find . -name "*.php" -exec grep -il "base64_decode\|eval" {} \;); do echo $file; head -n 3 $file; if [ wc -l < $file -gt 3 ]; then tail -n 3 $file fi; done | less

This returns the following error:

bash: syntax error near unexpected token `done'

Solution

I would to use the following

while read -r file
do
        echo  ==$file==
        head -n 3 "$file"
        [[ $(grep -c '' "$file") > 3 ]] && (echo ----last-3-lines--- ; tail -n 3 "$file")
done < <(find . -name \*.php -exec grep -il 'base64_decode\|eval' {} \+)

• Using while over the for is better, because the filenames could contain spaces. /probably not in this case, but anyway :)/
• using grep -c '' "$file" is sometimes better (when the last line in the file, doesn't contains the \n character (the wc counts the \n characters in the file)
• the find with the \+ instead of the \; is more efficient

Answered By - clt60

Answer Checked By - Mildred Charles (WPSolving Admin)