Pretty basic question, I'm trying to write a regex in Vim to match any phrase starting with "abc " directly followed by anything other than "defg".
I've used "[^defg]" to match any single character other than d, e, f or g.
My first instinct was to try /abc [^\(defg\)] or /abc [^\<defg\>] but neither one of those works.
Here's the search string.
/abc \(defg\)\@!
The concept you're looking for is called a negative look-ahead assertion. Try this in vim for more info:
:help \@!
preceeded or followed by?
If it's anything starting with 'abc ' that's not (immediately) followed by 'defg', you want bmdhacks' solution.
If it's anything starting with 'abc ' that's not (immediately) preceeded by 'defg', you want a negative lookbehind:
/\%(defg\)\@<!abc /
This will match any occurance of 'abc ' as long as it's not part of 'defgabc '. See :help \@<! for more details.
If you want to match 'abc ' as long as it's not part of 'defg.*abc ', then just add a .*:
/\%(defg.*\)\@<!abc /
Matching 'abc ' only on lines where 'defg' doesn't occur is similar:
/\%(defg.*\)\@<!abc \%(.*defg\)\@!/
Although, if you're just doing this for a substitution, you can make this easier by combining :v// and :s//
:%v/defg/s/abc /<whatever>/g
This will substitute '<whatever>' for 'abc ' on all lines that don't contain 'defg'. See :help :v for more.
Here we go, this is a hairy one:
/\%(\%(.\{-}\)\@<=XXXXXX\zs\)*
(replace XXXXXX with the search word). This will search for everything that does not contain XXXXXX. I imagine if you did:
/abc \%(\%(.\{-}\)\@<=defg\zs\)*
This may work like you want it to. Hope this helps a little!
I don't have enough reputation to comment Andy's answer, but his answer is wrong, so I explain it here, the expression"/abc\ [^d][^e][^f][^g]" is not match 'abc d111', 'abc def1' etc. the best way is the expression "/abc (defg)@!"