grep specific number from lines 5 but not 25 or 52 and so on

Question

I need to find a user for a userid. The return from the authentication system is as follows. Now with a bash script I need to extract the line with the exact number. Not 5 and 25 ..

------------
ID  LOGIN   
------------
28  user1
25  user2 
5   user3

If I use grep 5 I do get 2 lines but I need the line with "5".

Any ideas ?

Answer 1

There are several ways of doing it.

IMHO the best way is to use awk, which is useful when dealing with fields.

If you want a grep based solution, I would do:

grep -w '^5'

The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.

The awk solution would look like:

awk '$1 == 5'

If you want only the username, which is the second column:

awk '$1 == 5 {print $2}'

If you're searching for a string and not a numeric value, enclose the string in double quotes:

awk '$1 == "abc" {print $2}'

Answer 2

You could try with a regex (first char in line) and including the space:

grep -E "^5 "

Answer 3

Try the following:

query-auth-system | grep "^5\\>"

• ^: means "match at start of line
• \\>: matches a word-boundary. So it will match 5, but not 50.