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I want to pass values those are not equal zero (0). What is the best performing regex pattern for this ?
[^0]+
Means: Any character besides zero must occur at least once
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I need a regular expression that will find:
F.01,
F1.01,
F9.99,
F10.01,
And any decimal to the second place in between.
You can use this pattern
F\d+\.\d\d$
for most of the platforms it should work
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How to determine that string doesn't contain both symbols &# together using regular expression ?
You can use a negative lookahead:
/^(?!.*&#)(.*)/m
Demo
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Eg: only allow
< n digit>.<2 digit>
[0-9]{0,}\.[0-9]{2} if you need exactly 2 digit after the dot
[0-9]{0,}(\.[0-9]{1,2}){0,1} if precision can be empty
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In Perl, I would like to determine if a given string is valid, the rule is to check if values exist between commas.
e.g. abc,abc is a valid case, but abc, or abc,,abc are not.
m/^\s*,|,\s*,|,\s*$/
matches all invalid combinations, assuming whitespace does not count as "values".
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What is the a regular expression to only allow whole numbers from 1 through 99? (including both 1 and 99)
Try this:
^[1-9][0-9]?$
The ? means that the previous token is optional.
The ^ and $ are anchors for the start and end of the string respectively.
Surely you mean
[1-9][0-9]?