Shell Programming and Scripting

hi,

I am facing a problem when i use grep command in shell script.
problem is as follows...

1)i has declared an variable (say a)
2)Now i am searching for an pattern in a file and i used the followig command

cat /wls_../../scripts/log.txt | grep $a-JUL-08

i am not getting any output when i run the above command...

can some one please help out in sorting out the problem of how to pass an variable to grep command..

Thanks in advance.

Regards,
hemanth

swift response would be highly appreciated...

Worked on my HP-UX and CentOS box.

when you type: echo $a does it show you thats in $a?

You don't need to use the cat command (this is not the source of your problem) :

Code:

grep $a-JUL-08 /wls_../../scripts/log.txt

Execute the following commands and show us the output :

Code:

echo "a='$a'"
grep '$a-JUL-08' /wls_../../scripts/log.txt

Jean-Pierre.

Hi Ikon,

yes...when i type echo $a it prints the value thats in $a...

Regards,
hemanth

Hi aigles,

PFB the output of
echo "a='$a'"
grep '$a-JUL-08' /wls_../../scripts/log.txt

output:

a='29

here although log.txt contains data grep is not returning that data.

Regards,
hemanth

Looks like there's a newline at the end of the value of $a, would that explain it? Or did you just fail to copy over the closing single quote?