[SOLVED] gawk -F: '{if($1="kamlesh") print $0}' /etc/passwd

kamleshkk · 12-09-2015, 07:50 AM

gawk -F: '{if($1="<user-name>") print $0}' /etc/passwd
Please explain me the output,since it shows around 35-40 lines for a user of name user-name, but /etc/passwd file is having only single entry of that user.so from where all these lines come.

berndbausch · 12-09-2015, 08:07 AM

You need to use a double equal sign for comparisons:

Code:

{if($1=="<user-name>") print $0}

The way you do it, the following happens: You set $1 (the first field) to user-name, the if condition is true, and you print the entire line from the password file with the new user-name.

By the way, the awk way of doing things is as follows:

Code:

$1=="username" { print }

That is, you convert the if clause to a pattern.

thesnow · 12-09-2015, 08:15 AM

= is not the same as ==, so you are assigning a value not doing a comparison.

https://www.gnu.org/software/gawk/ma...Operators.html

kamleshkk · 12-09-2015, 08:26 AM

Quote:

Originally Posted by berndbausch

You need to use a double equal sign for comparisons:

Code:

{if($1=="<user-name>") print $0}

The way you do it, the following happens: You set $1 (the first field) to user-name, the if condition is true, and you print the entire line from the password file with the new user-name.

By the way, the awk way of doing things is as follows:

Code:

$1=="username" { print }

That is, you convert the if clause to a pattern.

Thank you very much, sometime even simple thing never comes in mind without help