Shell Programming and Scripting

Hi guys,

I was wondering if you could help me figure out what's wrong with this script below:

Code:

read username
 
if `grep $username /etc/passwd` > /dev/null
then
   $username >> information_file
else
   echo " "
fi

Apparently command is not found when script is run. Standard Output says the first "if" line is the culprit.

Many thanks!

This will attempt to run the value of $username as a command, which is not what you want to do. You probably want to do

Code:

   echo $username >> information_file

You should also validate that $username has a value before you try to use it in a grep.

Code:

 
   echo $username >> information_file

You should also validate that $username has a value before you try to use it in a grep.[/QUOTE]

Thanks for your reply man. Unfortuneatly, Bash still didn't accept the changes. What I want out of this if statement is to check if a username a user typed in can be found in etc/passwd. If yes, then to append that username variable to a text file called information_file.

You need to remove the backticks as well:

Code:

if grep $username /etc/passwd > /dev/null

Well, that made all the difference, thanks man!

Feel stupid now but when should you use back quotes then?

Back quotes turn the output of a program into a string. Use them to do that.

You should use them when you want to pass the output of what is within the quotes to another command, or a variable assignment.

e.g. if you wanted to assign the content of the user's passwd entry to a variable you would do

Code:

passwd_entry=`grep $username /etc/passwd`

The way you were using it, the output of the grep was being passed to the if statement, so the script tried to execute it as a command, which failed.