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06.13.2008 at 10:20AM PDT, ID: 23483460
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8.2

How to print output of grep

Asked by rae_rae in Project Management, Documentation in Programming

Tags:

I have attached a code snippet so you can see what I'm doing.

What I have is a directory filled with other directories and files containing .svn and .class.php files.

I need to come up with a way to determine if any of these files are being referenced in any of the other files.
This means I must grep through the files to find the string Filename.class.php but paths containing .svn should be ignored.

My comments indicate exactly my needs. But the code doesn't do exactly what they say. For instance,
set $file=$(ls -R1 |egrep -e '\.class\.php$') only gets the filename and not the directory/filename.

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#!/bin/bash
 
cd public_html/dd/
# Get the full path of the file
set $file=$(ls -R1 |egrep -e '\.class\.php$')
 
while [ $file ]
do
	# Ignore .svn directories 
	set $result=$(grep "$file" *.class\.php)
	# Only print resultant files not in a .svn directory
done
 
Keywords: How to print output of grep
Title
1 Grep multiple patterns
2 Help with grep & cut
3 Grep for multiple statements
4 Delete Files with grep?
5 grep
6 Grep for string in directories
 
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[+][-]06.13.2008 at 02:09PM PDT, ID: 21782818

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[+][-]06.14.2008 at 02:06AM PDT, ID: 21784648

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[+][-]06.25.2008 at 11:03AM PDT, ID: 21868261

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About this solution

Zones: Project Management, Documentation in Programming
Tags: Bash
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Solution Provided By: rae_rae
Participating Experts: 1
Solution Grade: A
 
 
 
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