Use the following:

/^\d*\.?\d*$/

• ^ - Beginning of the line;
• \d* - 0 or more digits;
• \.? - An optional dot (escaped, because in regex, . is a special character);
• \d* - 0 or more digits (the decimal part);
• $ - End of the line.

This allows for .5 decimal rather than requiring the leading zero, such as 0.5

/\d+\.?\d*/

One or more digits (\d+), optional period (\.?), zero or more digits (\d*).

Depending on your usage or regex engine you may need to add start/end line anchors:

/^\d+\.?\d*$/

Debuggex Demo

You need a regular expression like the following to do it properly:

/^[+-]?((\d+(\.\d*)?)|(\.\d+))$/

The same expression with whitespace, using the extended modifier (as supported by Perl):

/^  [+-]? ( (\d+ (\.\d*)?)  |  (\.\d+) ) $/x

or with comments:

/^           # Beginning of string
 [+-]?       # Optional plus or minus character
 (           # Followed by either:
   (           #   Start of first option
     \d+       #   One or more digits
     (\.\d*)?  #   Optionally followed by: one decimal point and zero or more digits
   )           #   End of first option
   |           # or
   (\.\d+)     #   One decimal point followed by one or more digits
 )           # End of grouping of the OR options
 $           # End of string (i.e. no extra characters remaining)
 /x          # Extended modifier (allows whitespace & comments in regular expression)

For example, it will match:

• 123
• 23.45
• 34.
• .45
• -123
• -273.15
• -42.
• -.45
• +516
• +9.8
• +2.
• +.5

And will reject these non-numbers:

• . (single decimal point)
• -. (negative decimal point)
• +. (plus decimal point)
• (empty string)

The simpler solutions can incorrectly reject valid numbers or match these non-numbers.

this matches all requirements:

^\d+(\.\d+)?$

Try this regex:

\d+\.?\d*

\d+ digits before optional decimal
.? optional decimal(optional due to the ? quantifier)
\d* optional digits after decimal

I ended up using the following:

^\d*\.?\d+$

This makes the following invalid:

.
3.

This is what I did. It's more strict than any of the above (and more correct than some):

^0$|^[1-9]\d*$|^\.\d+$|^0\.\d*$|^[1-9]\d*\.\d*$

Strings that passes:

0
0.
1
123
123.
123.4
.0
.0123
.123
0.123
1.234
12.34

Strings that fails:

.
00000
01
.0.
..
00.123
02.134

you can use this:

^\d+(\.\d)?\d*$

matches:
11
11.1
0.2

does not match:
.2
2.
2.6.9

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

should reflect what people usually think of as a well formed decimal number.

The digits before the decimal point can be either a single digit, in which case it can be from 0 to 9, or more than one digits, in which case it cannot start with a 0.

If there are any digits present before the decimal sign, then the decimal and the digits following it are optional. Otherwise, a decimal has to be present followed by at least one digit. Note that multiple trailing 0's are allowed after the decimal point.

grep -E '^[+-]?(([1-9][0-9]*)?[0-9](\.[0-9]*)?|\.[0-9]+)$'

correctly matches the following:

9
0
10
10.
0.
0.0
0.100
0.10
0.01
10.0
10.10
.0
.1
.00
.100
.001

as well as their signed equivalents, whereas it rejects the following:

.
00
01
00.0
01.3

and their signed equivalents, as well as the empty string.

What language? In Perl style: ^\d+(\.\d*)?$

What you asked is already answered so this is just an additional info for those who want only 2 decimal digits if optional decimal point is entered:

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

^ : start of the string
\d : a digit (equal to [0-9])
+ : one and unlimited times

Capturing Group (.\d{2})?
? : zero and one times . : character .
\d : a digit (equal to [0-9])
{2} : exactly 2 times
$ : end of the string

1 : match
123 : match
123.00 : match
123. : no match
123.. : no match
123.0 : no match
123.000 : no match
123.00.00 : no match

try this. ^[0-9]\d{0,9}(\.\d{1,3})?%?$ it is tested and worked for me.

Regular expression:

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

use \d+ instead of \d{0,1} if you want to allow more then one number use \d{0,2} instead of \d{0,1} if you want to allow up to two numbers after coma. See the example below for reference:

or

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

or

^\d+((.)|(.\d+)?)$

Explanation

(These are generated by regex101)

• ^ asserts position at start of a line
• \d matches a digit (equivalent to [0-9])
• + matches the previous token between one and unlimited times, as many times as possible, giving back as needed (greedy)
• 1st Capturing Group ((.)|(.\d{0,1})?)
• 1st Alternative (.)
• 2nd Capturing Group (.)
• . matches any character (except for line terminators)
• 2nd Alternative (.\d{0,1})?
• 3rd Capturing Group (.\d{0,1})?
• ? matches the previous token between zero and one times, as many times as possible, giving back as needed (greedy)
• . matches any character (except for line terminators)
• \d matches a digit (equivalent to [0-9])
• {0,1} matches the previous token between zero and one times, as many times as possible, giving back as needed (greedy)
• $ asserts position at the end of a line