Bash command - using IF - FI within a DO - DONE - bash

Question

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'

Answer 1

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

Answer 2

Problem seems to be here:
if [ wc -l < $file -gt 3 ]; then
Since you need to use command substitution here to make sure wc -l command executes first and then compare the result:
if [[ $(wc -l < "$file") -gt 3 ]]; then

Answer 3

You want to execute your wc, more like:
if [[ $(wc -l < $file) -gt 3 ]]; then

Answer 4

try this:
#!/bin/bash
for file in $(grep -H "base64_decode\|eval" ./*.php | cut -d: -f1);
do
echo $file;
head -n 3 $file;
if [[ $(wc -l < $file) -gt 3 ]];
then
tail -n 3 $file
fi;
done
I tested and seems to work fine.
But, be carefull ... if php has 4 lines, you will see:
line1
line2
line3
line2
line3
line4
EDIT: changed the script above to grep inside files.
cat a.php
asdasd
asd
base64_decode
l
a
and result
./test2.sh
./a.php
asdasd
asd
base64_decode
base64_decode
l
a