Preg_match PHP -- pattern matching

Question

trying to understand preg_match, struggling to understand how to write and how to access what it has matched. For example:

Every single movie name I have is in the format--

MOVIE NAME (YEAR)

e.g. Alice in Wonderland (2010)

I want to be able to get the movie title into a different variable from the string.

A few movies have parentheses outside of the year -- as in, The movie (Has One of These) (2008)

I'm iterating over an array of strings as well -- so I basically need to use preg_match to get to $[0-9]{4}$$ (is $ the mark of an end of the line?) and then the rest of the string without that year as well in two variables.

Can anyone possibly help?

EDIT: Huh. I swear I typed \ . When I type \( it went into ( because I didn't double escape. Anyway, thank you guys very much! The site you linked it also awesome (helped with array problems, I didn't realize it kept full string at 0 as well).

It seems like you're almost there. What is the problem you're having exactly, though? What do you want help with exactly? — pgl
– pgl, Commented Dec 8, 2011 at 10:20
I guess I really don't understand how to skip over everything at the beginning in terms of regex terms — Paul
– Paul, Commented Dec 8, 2011 at 10:23
Well, you kind of have with the regex you used - the only difference is that you'd need to escape the ()s. ie: ([0-9]{4})$ should match the (nnnn) year at the end of a movie title. — pgl
– pgl, Commented Dec 8, 2011 at 10:35
@Paul did any of the answers helped you, if so. upvote them and check as correct the answer you think was more helpful. — SERPRO
– SERPRO, Commented Dec 8, 2011 at 11:00

SERPRO · Accepted Answer · 2011-12-08 10:57:11Z

3

well if your pattern is: SOMETHING + (YEAR) then your regex should be like this:

 #^(.+)\((\d{4})\)$#

Explanation:

 # -> pattern delimiter
 ^ -> beginning of string
 (.+) -> any character "." once or more "+"
 \( -> escape parenthesis character
 \d{4} -> four digits
 \) -> escape parenthesis character
 $ -> end of string

Example

edited Dec 8, 2011 at 10:57

answered Dec 8, 2011 at 10:24

SERPRO

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Comments

hakre · Accepted Answer · 2011-12-08 10:25:37Z

1

Looking for patterns in these lines:

Alice in Wonderland (2010)
The movie (Has One of These) (2008)

You suggested in your question to use the following regular expression:

([0-9]{4})$

to match the year at the end of the line. $ is infact a marker for the end of the line, however, the ) is a special character in a regular expression that needs to be slashed to work:

\(([0-9]{4})\)$
  ^         ^^ both brackets have been slashed to match them literally
  '- subgroup parenthesis.

or by using \d for any decimal number:

\((\d{4})\)$

This will make subgroup 1 contain the year.

answered Dec 8, 2011 at 10:25

hakre

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KuKu · Accepted Answer · 2011-12-08 10:31:34Z

0

This is one of the link where regex man be made

You also can use following for decimal number

\((\d{4})\)$

answered Dec 8, 2011 at 10:31

KuKu

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Snifff · Accepted Answer · 2011-12-08 11:02:33Z

0

Here's an online regexp tester, it is very useful when learning regular expressions and testing them: HiFi Regex Tester.

answered Dec 8, 2011 at 11:02

Snifff

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