Regex how to match an optional character

Question

I have a regex that I thought was working correctly until now  I need to match on an optional character  It may be there or it may not   Here are two strings  The top string is matched while the lower is not  The absence of a single letter in the lower string is what is making it fail   I d like to get the single letter after the starting 5 digits if it s there and if not  continue getting the rest of the string  This letter can be A-Z   If I remove   A-Z  1          from the regex  it will match everything I need except the letter but it s kind of important   20000      K               Q511195DREWBT            E00078748521 30000                      K601220PLOPOH            Z00054878524   Here is the regex I m using       0-9  5         A-Z  1           A-Z  1    0-9  3    0-9  3    A-Z  3    A-Z  3      A-Z   0-9  3   0-9  4    0-9  2    0-9  2

User · Accepted Answer

Use   A-Z     to make the letter optional   1  is redundant   Of course you could also write  A-Z  0 1  which would mean the same  but that s what the   is there for    You could improve your regex to     0-9  5    s   A-Z    s   A-Z    0-9  3    0-9  3    A-Z  3    A-Z  3   s   A-Z   0-9  3   0-9  4    0-9  2    0-9  2     And  since in most regex dialects   d is the same as  0-9       d 5    s   A-Z    s   A-Z    d 3    d 3    A-Z  3    A-Z  3   s   A-Z   d 3   d 4    d 2    d 2     But  do you really need 11 separate capturing groups  And if so  why don t you capture the fourth-to-last group of digits

User · Answer

You also could use simpler regex designed for your case like              n r     where  2 match what you want

User · Answer

You can make the single letter optional by adding a   after it as     A-Z  1      The quantifier  1  is redundant so you can drop it

User · Answer

You have to mark the single letter as optional too     A-Z  1             or make the whole part optional     A-Z  1

[regex] Regex how to match an optional character

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