Simple regular expression for a decimal with a precision of 2

Question

What is the regular expression for a decimal with a precision of 2?

Valid examples:

123.12
2
56754
92929292929292.12
0.21
3.1

Answer 1

This will allow decimal with exponentiation and upto 2 digits ,

^[+-]?\d+(\.\d{2}([eE](-[1-9]([0-9]*)?|[+]?\d+))?)?$

Demo

Answer 2

Valid regex tokens vary by implementation. A generic form is:

[0-9]+(\.[0-9][0-9]?)?

More compact:

\d+(\.\d{1,2})?

Both assume that both have at least one digit before and one after the decimal place.

To require that the whole string is a number of this form, wrap the expression in start and end tags such as (in Perl's form):

^\d+(\.\d{1,2})?$

To match numbers without a leading digit before the decimal (.12) and whole numbers having a trailing period (12.) while excluding input of a single period (.), try the following:

^(\d+(\.\d{0,2})?|\.?\d{1,2})$

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Added

Wrapped the fractional portion in ()? to make it optional. Be aware that this excludes forms such as 12. Including that would be more like ^\d+\\.?\d{0,2}$.

Added

Use ^\d{1,6}(\.\d{1,2})?$ to stop repetition and give a restriction to whole part of the decimal value.

Answer 3

preg_match("/^-?\d+[\.]?\d\d$/", $sum)

Answer 4

For numbers that don't have a thousands separator, I like this simple, compact regex:

\d+(\.\d{2})?|\.\d{2}

or, to not be limited to a precision of 2:

\d+(\.\d*)?|\.\d+

The latter matches
1
100
100.
100.74
100.7
0.7
.7
.72

And it doesn't match empty string (like \d*.?\d* would)

Answer 5

In general, i.e. unlimited decimal places:

^-?(([1-9]\d*)|0)(.0*[1-9](0*[1-9])*)?$.