What exactly does .*? do in regex? ".*?([a
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The ? here acts as a 'modifier' if I can call it like that and makes .* match the least possible match (termed 'lazy') until the next match in the pattern.
In fall/2005 , the first .*? will match up to the first match in ([am/]*) , which is just before f . Hence, .*? matches 0 characters so that ([am/]*) will match fall/ and since ([am/]*) cannot match anymore, the next part of the pattern .* matches what's left in the string, meaning 2005 .
In contrast to .*([am/]*).* , you would have .* match as much as possible first (meaning the whole string) and try to go back to make the other terms match. Except that the problem is with the other quantifiers being able to match 0 characters as well, so that .* alone will match the whole string (termed 'greedy').
Maybe a different example will help.
.*ab
In:
aaababaaabab
Here, .* will match as much characters as possible and then try to match ab . Thus, .* will match aaababaaab and the remainder will be matched by ab .
.*?ab
In:
aaababaaabab
Here, .*? will match as little as possible until it can match ab in that regex. The first occurrence of ab is here:
aaababaaabab
^^
And so, .*? matches aa while ab will match ab .
In regex:
? : Occurs no or one times, ? is short for {0,1}
* ? : ? after a quantifier makes it a reluctant quantifier, it tries to find the smallest match.
Suppose if you have a string input like this
this is stackoverflow
and you use regex
.*
so output will be
this is stackoverflow
but if you use regex
.*?
your out put will be
this
So from the above example it is clear that if you use .* it will give you whole string. to prevent this if you want only first cherector before space you should use .*?
For more practical knowledge you can check http://regexpal.com/
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