[regex] Regular Expression with wildcards to match any character

Without knowing the exact regex implementation you're making use of, I can only give general advice. (The syntax I will be perl as that's what I know, some languages will require tweaking)

Looking at ABC: (z) jan 02 1999 \n

  • The first thing to match is ABC: So using our regex is /ABC:/

  • You say ABC is always at the start of the string so /^ABC/ will ensure that ABC is at the start of the string.

  • You can match spaces with the \s (note the case) directive. With all directives you can match one or more with + (or 0 or more with *)

  • You need to escape the usage of ( and ) as it's a reserved character. so \(\)

  • You can match any non space or newline character with .

  • You can match anything at all with .* but you need to be careful you're not too greedy and capture everything.

So in order to capture what you've asked. I would use /^ABC:\s*\(.+?\)\s*(.+)$/

Which I read as:

Begins with ABC:

May have some spaces

has (

has some characters

has )

may have some spaces

then capture everything until the end of the line (which is $).

I highly recommend keeping a copy of the following laying about http://www.cheatography.com/davechild/cheat-sheets/regular-expressions/