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I need a regular expression pattern that matches any number including 1-9 numbers except 2?
My attempt:
([1-9][^2])
But this doesn't work for me.
702
To match any number from 0 to 9 we use \d in regex. It will match any single digit number from 0 to 9. \d means [0-9] or match any number from 0 to 9. Instead of writing 0123456789 the shorthand version is [0-9] where [] is used for character range.
$ means "Match the end of the string" (the position after the last character in the string).
In a regular expression, if you have [a-z] then it matches any lowercase letter. [0-9] matches any digit. So if you have [a-z0-9], then it matches any lowercase letter or digit.
Example: Regex Number Range 1-20 Range 1-20 has both single digit numbers (1-9) and two digit numbers (10-20). For double digit numbers we have to split the group in two 10-19 (or in regex: "1[0-9]") and 20. Then we can join all these with an alternation operator to get "([1-9]|1[0-9]|20)".
Another way to do it:
/[^\D2]/
Which means, not a non-digit or 2.
81
You can match the range of numbers before and after two with [0-13-9], like this:
"4526".match(/[0-13-9]+/)
["45"]
"029".match(/[0-13-9]+/)
["0"]
"09218".match(/[0-13-9]+/)
["09"]
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