For example I have this RegEx:
([0-9]{1,4})([0-9])
Which gives me these matching groups when testing with string "3041":
As you can see, group2 is filled before group1 even if the quantifier is greedy.
How can I instead make sure to fill group1 before group2?
EDIT1: I want to have the same regEx, but have "3041" in group1 and group2 empty.
EDIT2: I want to have "3041" in group1 and group2 empty. And, yes, I want the regEx to not match!,
For an input "1234", the pattern: ([0-9]{1,4})([0-9]) is being as greedy as possible.
The first capture group cannot contain four characters, otherwise the last part of the pattern would not match.
Perhaps what you're looking for is:
([0-9]{1,4})([0-9]?)
By making the second group optionally empty, the first group can contain all four characters.
Edit:
I want the regEx to not match!, I want only 5 digits strings to match the whole RegEx.
In this case, your pattern should not really be "1-4 characters" in the first group, since you only want to match a group of 4:
([0-9]{4})([0-9])
In some regex flavours (i.e. not all languages support this), it is also possible to make quantifiers possessive (although this is unnecessary in your case, as shown above). For example:
([0-9]{1,4}+)([0-9])
This will force the first group to match as far as it can (i.e. 4 characters), so a 3-character match does not get attempted and the overall pattern fails to match.
Edit2:
Is "possessiveness" available in Javascript? If not, any workarounds?
Unfortunately, possessive quantifiers are not available in JavaScript.
However, you can emulate the behaviour (in a slightly ugly way) with a lookahead:
(?=([0-9]{1,4}))\1([0-9])
In general, a possessive quantifier a++ can be emulated as: (?=(a+))\1.
[0-9]is an obligatory pattern. If you make the first one possessive, this whole pattern will never match any string. - Wiktor Stribiżew([0-9]{1,4})([0-9]?)- Wiktor Stribiżew