Regex: Everything except some pattern

Question

I have a string:

foo bar
foo1 #9 0x103806f4 bar1
foo2 #10 0x0f6dd704 bar2
foo3 bar3

I have tried the following:

^((?!#[\\d]{1,2} 0x[0-9a-f]{8}).)*$

which gets

foo bar
foo3 bar3

and

^((?!#[\\d]{1,2} 0x[0-9a-f]{8}).)*

which gets

foo bar
foo1
foo2
foo3 bar3

But what im trying to get is

foo bar
foo1 bar1
foo2 bar2
foo3 bar3

How can I achieve this?

Answer 1

You need to do replace instead of matching in-order to get the desired output.

\s*#\d{1,2} 0x[0-9a-f]{8}

Use the above regex and then replace the match with an empty string.

DEMO

Answer 2

If you're wanting the beginning and ending non-whitespace characters, using a Negative Lookahead is not going to do the job. You could match your expected output as follows:

^(\S+).*?(\S+)$

Then in your preferred language, you can combine the match results: python example ...

>>> import re
>>> s = '''foo bar
foo1 #9 0x103806f4 bar1
foo2 #10 0x0f6dd704 bar2
foo3 bar3'''
...
>>> for m in re.finditer(r'(?m)^(\S+).*?(\S+)$', s):
...     print(" ".join(m.groups()))

foo bar
foo1 bar1
foo2 bar2
foo3 bar3

Instead of using regex, consider splitting the string and joining the indexes together.