Shell Programming and Scripting

I need to grep for pattern "enable_cal = true".The problem is that different file has different no of spaces in that string.

for eg

one file will have "enable_cal <space><space><space>=true"
next file will have "enable_cal= <space><space>true"
other one will have "enable_ca<space><space>=<space><space>true"

Now I am seperately grep for each pattern.Plz help me here, I want a single grep that would only look for "enable_cal=true" provided how many spaces included between each word.

is this what you want??

Code:

grep "enable_cal.*=.*true" filename

Thanks dude.The code is working fine.
Thanks once again.

Just the continuation to where I left.

I just want to use grep to search for files that doesn't have 2 specific patterns "enable_cal=true" and "enable_cal=false"

I used the command below( grep -lv) is that correct

find ./ -name "cal_client.txt" | xargs grep -lv `enable_cal.*=.*false|enable_cal.*=.*true` > temp6.txt


But I am not getiing expected o/p.The expected output is 7 files only.But somehow I am getting 229 files.

Added details
==========
234 files total
202 files with pattern "enable_cal=false"
25 files with pattern "enable_cal=true"

I just want the pending 7 file names that doesn't has both the patterns.

remove v from -lv because -v will return the names of the file in which pattern is not found.. refer grep man page

If I remove -v option(with still -l option), I am getting an empty file.

Plz help me.