Shell Programming and Scripting

Hello, I have an input file that looks like so:

Code:

LDLR
LDLRAD4
VLDLR

when I grep "LDLR" I get an output of:

Code:

LDLR
LDLRAD4
VLDLR

Since all names have "LDLR" included within them, but all I want the output to be is

Code:

LDLR

I know it can work if I surround the words with pipes for instance and grep"|LDLR|", but my actual input file is thousands of lines long.

Any help is appreciated!

Try this with a modern awk, not solaris awk:

Code:

 awk '/\<LDLR\>/' filename

Is there always only one "word" per input line? Or, are there one or more "words" separated by spaces, tabs, or punctuation marks? If there can be multiple words on a line, do you want the input line containing the word LDLR printed, or do you want to print entire lines when the word LDLR is one of the words on the line?

If there is one word per line and you want to print each line that just contains that word, the simple solution is:

Code:

grep '^LDLR$' input_file

if your file is space delimited, you can try

Code:

grep -w LDLR

Note that neither the -w option for grep nor the \<ERE\> in an awk ERE are in the standards. They will work in some awk and grep utilities, but they are not portable.

Code:

fgrep -x LDLR

NB fgrep takes plain strings.