[SOLVED] grep & regex

[FearlessPc]

Hello

I'm trying to look at all the processes excluding "root"


ps -ef | grep '[^root]'

and "root" always shows in the output
I tried grep with -G and -E with no luck

also tried

grep '[^root]' /etc/passwd
no luck

What I'm doing wrong ?

[mister_pink]

Excluding root? For a start you want the -v option then - to get lines that do not match. Then lose the square brackets.

ps -ef | grep -v '^root'

[FearlessPc]

Thanks working great.

But why [^x] didn't work ?
all the guides i looked in (3) says "The ^ inside square brackets negates the expression".

[mister_pink]

I'm by no means an expert, but the only use I know of for a ^ is as an anchor - meaning what follows must start the line. Square brackets create a character set, so for example [root] matches the letters 'r', 'o' or 't'. I suspect that in this context ^ reverses that, meaning [^root] matches single character except r o and t.

Again though, not entirely certain. Theres lots of sites about regex though

[mbsullivan]

know of for a ^ is as an anchor - meaning what follows must start the line. Square brackets create a character set, so for example [root] matches the letters 'r', 'o' or 't'. I suspect that in this context ^ reverses that, meaning [^root] matches single character except r o and t.

This is correct. Therefore, if you wanted a grep that will be equivalent to the version that works:


ps -ef | grep -v '^root'

You would have to use something like the following:


ps -ef | grep '^[^r][^o][^o][^t]'

This says: "print lines that begin with any four characters except an r followed by an o followed by an o followed by a t".

You can use () to group characters, such that this should also work:


ps -ef | grep '^[^(root)]'

Due to its simplicity, I think the -v option is preferred.

Mike

[FearlessPc]

Thank you both for the solutions & clarification, my mistake was not grouping "root" together and the -v option is much more simpler in this case.