I am searching for a regular expression that matches any string that is:
- A number which is greater than zero
- The number has at least one digit
- No more than 12 digits
I tried this one without success:
^[1-9][0-9]*{1,12}$ 
- Can you have numbers that contain leading zeros but are greater than zero?Sebastian Proske– Sebastian Proske2016-10-27 07:53:53 +00:00Commented Oct 27, 2016 at 7:53
- @SebastianProske Would be nice but does not have to be like that. Interesting point!NoobieNoob– NoobieNoob2016-10-27 08:09:29 +00:00Commented Oct 27, 2016 at 8:09
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4 Answers
If numbers with leading zeros that are greater than zero are allowed, you can use ^(?!0+$)[0-9]{1,12}$ if the tool/language you use supports lookaheads. The lookahead is used to ensure that the number doesn't entirely consist of zeros.
Comments
The problem with the regular expression
^[1-9][0-9]*{1,12}$ is the star * used together with {1,12} as the star *means any number of repetitions of the previous symbol whereas {1,12} means one to twelve repetitions - so the star * and {a,b} don't mix and must be used exclusively, not together.
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Use ^[1-9]\d{0,11}$, which checks for a non-zero digit followed by zero to 11 digits.
Your regex had a quantifier * after the second character class, which would allow zero or more occurrence of a digit (which can be more than 11 digits). Also the quantifier {1,12} sets the minimum occurrence of preceding pattern as 1 and maximum occurrence as 12. So you had to avoid the * and set minimum and maximum as 0 and 11 respectively, since there was already a pattern for single digit.
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^[1-9][0-9]{0,11}$
Start with a single digit between 1-9, then have 0 to 11 occurrences of a digit between 0-9
2 Comments
CodeMonkey
Why do you have a capture group around the last 0-11 digits?
Vasan
You're right, my mistake. I was testing something else with this expression and forgot to remove the group. Updated now!