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Question

Answer 1

+2 A:

Try this:

x = "XJ FC-X35 (1492)"
x.replace(/\s*\(.*?\)/,'');

Andrew 2010-09-21 16:30:01

Answer 2

+4 A:

replace(/^(.*?)\s?\(([^)]+)\)$/gi, "$1$2")

Takes XJ FC-X35 (1492) and returns XJ FC-X351492.

Remove the $2 to turn XJ FC-X35 (1492) into XJ FC-X35, if that's what you wanted instead.

Long explanation

^   // From the start of the string
(   // Capture a group ($1)
.*? // That contains 0 or more elements (non-greedy)
)   // Finish group $1
\s? // Followed by 0 or 1 whitespace characters
\(  // Followed by a "("
(   // Capture a group ($2)
[   // That contains any characters in the following set
^)  // Not a ")"
]+  // One or more times
)   // Finish group $2
\)$ // Followed by a ")" followed by the end of the string.

Sean Vieira 2010-09-21 16:30:52

+1 for a robust regexp, and for providing alternatives depending on what the OP meant. But do you really need to match and replace the initial `^(.*?)`? Why not just omit it?

LarsH 2010-09-21 16:36:25

@LarsH - Habit more than anything else. I always prefer, when dealing with regular expressions to make sure that I'm grabbing what I want, rather than removing what I don't (whitelist, don't blacklist). You are correct though - the `^(.*?)` and `$` could be removed without harm resulting in the easier to read `replace(/\s?$([^)]+)$/gi, "$1")` -- and it's up to the fair reader to choose the better way.

Sean Vieira 2010-09-21 16:41:17

@Sean - nice bit-by-bit explanation. Btw I preferred your earlier `\s*` over `\s?`, since the OP's "the whitespace" is pretty ambiguous as to cardinality. I would have put `\s+` but you may be right.

LarsH 2010-09-21 17:00:01

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