UNIX for Dummies Questions & Answers

Code:

cat testfile.txt | grep -w '^root'

or

Code:

cat testfile.txt | grep '^root'

i tried both the ways, no output for the above commands.

The only reason for not getting any output is that there are no records that starts with root

Check your input file carefully for any blank spaces or other characters.

I get record for the below command.

Code:

cat testfile.txt  | grep 'root'

The moment i add the special character "^", no records showing out.

---------- Post updated at 01:22 PM ---------- Previous update was at 01:01 PM ----------

I tried like below, it worked.

Code:

cat testfile.txt  | awk '$1 == "root"'

but how to use shell variable instead of literal name in the place of "root".

See post #2

And please no UUOC!
Other variants:

Code:

<testfile.txt awk '$1 == "'"$var"'"'

Better readable is

Code:

awk '$1==key' key="$var" testfile.txt

Code:

perl -slane 'print if $F[0] eq $pat' < file -- -pat='root'

The reason is most probably that "root" is the first field, but not the first characters on the line:

Code:

root     blabla .....
 root    blabla .....

The expression will find the first line, but not the second, because of the leading blank. "^" means "begin of line" and "^root" means "begin of line, immediately followed by r-o-o-t". This is the case in the first example, not the second.

If you want to allow for optional blanks at the beginning of line modify the regular expression this way ("<b>" denotes a literal blank, "<t>" a tab character):

Code:

grep '^[<b><t>]*root' /some/file'

This means "begin of line, followed by zero or more space- or tab-characters (=white space), followed by the string r-o-o-t". This would find both lines in the example above.

I hope this helps.

bakunin