Shell Programming and Scripting

Hi
I have a requirement like

Code:

# e=`ls | grep -e"test" | awk -F " " '{print $1}'`

(0) root @ ptxfv3: /
# echo $e
test test.xml

From this i need to grep the word "test" alone i.e., it is occuring twice I need only one among them

Please help

The output is like this,

ls ---> list all files & directories from the current directory

grep -e test ---> filter all files/directories with the name having "test" in it

Code:

NOTE: this output will be line-by-line
Ex:
test
test.xml

awk -F " " '{print $1}' ---> This will push all the matched files to your variable "e"

I would suggest to use

Code:

e=$(ls | grep -e "test" | head -1)

How about using anchors?

Code:

e=$(ls | grep "^test$")

Is the requirement for grepping the word "test" ALONE

or

to grep the first occurrence of the word "test" ???

Your requirements are extremely ambiguous. If I take them literally, I interpret it to mean that if I have files named attest, bestest, and test.c (or, in fact, one or more files with names containing the string "test" even if none of them are actually named "test"); you want to print test. And, if there are no filenames in the current directory that contain the string "test", you don't want to print anything. If that is what you want, the following will do that if you're using a recent bash or a Korn shell newer than 1988:

Code:

list=( *test* )
if [ ${#list[@]} -gt 1 ] || [ "${list[0]}" != "*test*" ]
then    echo test
fi

If that isn't what you want. Please clearly state what you are trying to do.

Code:

$ ls test*
test  test~
$ ls | awk '/test/ {print $1; exit}'
test

Its the first occurence of the word test