Decimal number regular expression where digit after decimal is optional

Question

I need a regular expression that validates a number  but doesn t require a digit after the decimal  ie   123 123  123 4   would all be valid  123     would be invalid  Any would be greatly appreciated

User · Answer

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

User · Answer

I ended up using the following     d     d     This makes the following invalid     3

User · Answer

Try this regex    d     d        d   digits before optional decimal       optional decimal optional due to the   quantifier     d   optional digits after decimal

User · Answer

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

User · Answer

In Perl  use Regexp  Common which will allow you to assemble a finely-tuned regular expression for your particular number format  If you are not using Perl  the generated regular expression can still typically be used by other languages   Printing the result of generating the example regular expressions in Regexp  Common  Number     perl -MRegexp  Common number -E  say  RE num  int          -        0123456789        perl -MRegexp  Common number -E  say  RE num  real        i     -               0123456789      0123456789                 0123456789  0             E         -        0123456789           perl -MRegexp  Common number -E  say  RE num  real  -base  gt 16        i     -               0123456789ABCDEF      0123456789ABCDEF                 0123456789ABCDEF  0             G         -        0123456789ABCDEF

User · Answer

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

User · Answer

you can use this     d     d   d     matches  11 11 1 0 2    does not match   2 2  2 6 9

User · Answer

lt    d   d       d        d     clean and simple   This uses Suffix and Prefix  RegEx features   It directly returns true - false for IsMatch condition

User · Answer

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  amp  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

User · Answer

What language  In Perl style    d     d

User · Answer

d         d        Came up with this  Allows both integer and decimal  but forces a complete decimal  leading and trailing numbers  if you decide to enter a decimal

User · Answer

-     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

User · Answer

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

User · Answer

try this    0-9  d 0 9     d 1 3       it is tested and worked for me

User · Answer

I think this is the best one because it matches all requirements    d      d

[regex] Decimal number regular expression, where digit after decimal is optional

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