Shell Programming and Scripting

You must allow some characters in between

Code:

ls -l | grep '^[l].*'"$argv[1]"'$'

Or shorter

Code:

ls -l | grep '^l.*'"$1"'$'

You need to allow some characters in between.

Code:

 ls -l | grep '^l.*softlink.txt'

The "." matches any single character and "*" means that the preceding item, any character, will be matched zero or more times.

sammythesp3rmy,
The argument you passed to grep will only match the line:

Code:

l "$argv1"

There is a big difference between single quotes and double quotes, and double quotes inside single quotes are taken as literal characters.

I got it to work by using MadeInGermany's suggestion. Would you be able to explain what happened so I can understand? What do the second pair of single quotes around "$argv[1]" actually help it do to work? When you say the single quotes around that help make it a literal character, that the single quotes actually make what's inside the parenthesis an actual string, while argv[1] gets put into softlink.txt?

Not around the "$argv[1]".
For the shell it is a concatenation of three strings

Code:

'string1'"string2"'string3'