I want to search the file /etc/passwd for all lines containing 'csh' but exlude all those lines that have '/usr' in them and dump the results into the file result.

IMPORTANT: I need to do this in one command line.

The following does not work:

grep -v \(\/usr\) \(csh\) /etc/passwd > result

What is the basic thing that I am doing incorrectly here?

do I need do first search for all instances of 'csh' and then direct this into another grep for exclude all instances of '/usr' ??

grep -v '/usr' /etc/passwd | grep csh > result


regards and thanks,

Steve

Why does this have to be on one command line? This sounds like a homework exercise - and that is not what this forum is for.......(just so you know).

The suggestion you made will work.....have you tried it?

grep -v '/usr' /etc/passwd | grep csh > result

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Pete

It's a homework question...
I thought I'd responded to a similar question a little while ago, see here .

yeah, I did figure this out on my own... that is why I put in the correct answer when I submitted the question.

I wasn't clear as to my real question or state of confusion...

I was struck by how simplistic these commands strings must be, and thought I was missing something.

When I tried to figure out the answer I just thought that there might be a way to put multiple arguments in one grep command instead passing the output onto another grep command.

Many years ago I did BLISS programming and I just have a lot of cobwebs stuck in my brain!

Well you could always write a new grep command..... that does what I suggested does...

new-grep /usr csh > result.

Some more homework!

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Pete