Shell Programming and Scripting

hey guys,

I have a directory with about 600 files. I need to find a specific word inside a command and replace only that instance of the word in many files. For example, lets say I have a command called 'foo' in many files. One of the input arguments of the 'foo' call is 'bar'. The word 'bar' many occur in many instances in a file but I want to replace only the 'bar' string inside the foo command call with bar1. I tried the following:

Code:

find . -type f -name *.c  | grep -l foo | xargs perl -ie "s/bar/bar1/g"

This finds all files with the foo command call but replaces all instances of bar with bar1 in those files. I am not sure how to make more specific and make it replace only the foo(bar) instance.

Thanks in advance !

You've almost got it.

Code:

find . -type f -name *.c | grep -l foo | xargs perl -ie "s/foo\(bar\)/foo\(bar1\)"

or something like that. Figure out what uniquely identifies your foo command call and add that into the first part of the search and replace.

"grep -l foo " is wrong, it greps on filename, not the file content.

Get the file list with foo in content:

Code:

find . -type f -name "*.c" -exec grep -l foo {} \;

rdcwayx is absolutely correct. Good catch!

Hi,

Try this,

Code:

perl -e 'while(<>){ s/(?<=foo)bar/bar1/; print $_; }' infile

inputfile

Code:

test foobar
test1 bar
testbar
test3bar
test5 foobar

output

Code:

test foobar1
test1 bar
testbar
test3bar
test5 foobar1

Yea .. good catch .. I forgot to type 'xargs' before grep ..

Code:

find . -type f -name *.c  | xargs grep -l foo

This did the right thing and found all the relevant files.

I probably did not mention that the 'bar' can appear anywhere in the foo command call and there is no specific pattern. Will I be able use wildcards in the pattern matching string for 's/' and say inside the foo command change all occurrences of bar? I tried using foo\(.*)bar pattern but i keep getting the signal 13 error from grep or unmatched ) error. I am surely missing something obvious here. pointers?

---------- Post updated at 03:26 PM ---------- Previous update was at 02:57 PM ----------

I went a step further now ... I tried the following:

Code:

perl -ie 's/foo.*?bar/foo.*?bar1/'

It matched the proper pattern but instead of replacing the existing contents and only replacing bar with bar1, it replaced the whole thing with 'foo.*?bar1' which is not what I wanted.

Code:

foo(blah,blah,blah,bar,blah,blah)

should be changed to

Code:

foo(blah,blah,blah,bar1,blah,blah)

The number of blah's varies widely. How do I make it replace the pattern properly?