Regular Expression with wildcards to match any character

Question

I am new to regex and I am trying to come up with something that will match a text like below:

ABC: (z) jan 02 1999 \n

Notes:

• text will always begin with "ABC:"
• there may be zero, one or more spaces between ':' and (z).
• Variations of (z) also possible - (zz), (zzzzzz).. etc but always a non-digit character enclosed in "()"
• there may be zero,one or more spaces between (z) and jan
• jan could be jan, january, etc
• date couldbe in any format and may/may not contain other text as part of it so I would really like to know if there is a regex I can use to capture anything and everything that is found between '(z)' and '\n'

Any help is greatly appreciated! Thank you

Answer 1

The following should work:

ABC: *\([a-zA-Z]+\) *(.+) 

Explanation:

ABC:            # match literal characters 'ABC:'  *              # zero or more spaces \([a-zA-Z]+\)   # one or more letters inside of parentheses  *              # zero or more spaces (.+)            # capture one or more of any character (except newlines) 

To get your desired grouping based on the comments below, you can use the following:

(ABC:) *(\([a-zA-Z]+\).+) 

Answer 2

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/