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
Invalid examples:
12.1232
2.23332
e666.76
The decimal point may be optional, and integers may also be included.
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
Invalid examples:
12.1232
2.23332
e666.76
The decimal point may be optional, and integers may also be included.
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})$
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}$.
Use ^\d{1,6}(\.\d{1,2})?$ to stop repetition and give a restriction to whole part of the decimal value.
^[0-9]+(\.[0-9]{1,2})?$
And since regular expressions are horrible to read, much less understand, here is the verbose equivalent:
^ # Start of string
[0-9]+ # Require one or more numbers
( # Begin optional group
\. # Point must be escaped or it is treated as "any character"
[0-9]{1,2} # One or two numbers
)? # End group--signify that it's optional with "?"
$ # End of string
You can replace [0-9] with \d in most regular expression implementations (including PCRE, the most common). I've left it as [0-9] as I think it's easier to read.
Also, here is the simple Python script I used to check it:
import re
deci_num_checker = re.compile(r"""^[0-9]+(\.[0-9]{1,2})?$""")
valid = ["123.12", "2", "56754", "92929292929292.12", "0.21", "3.1"]
invalid = ["12.1232", "2.23332", "e666.76"]
assert len([deci_num_checker.match(x) != None for x in valid]) == len(valid)
assert [deci_num_checker.match(x) == None for x in invalid].count(False) == 0
This will allow decimal with exponentiation and upto 2 digits ,
^[+-]?\d+(\.\d{2}([eE](-[1-9]([0-9]*)?|[+]?\d+))?)?$