Validate decimal greater than 1 with regex

user_14628299 0
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I want to validate that a number with three optional integer and four digits mandatory after the decimal point is greater than 1. For example:

0.0000     [No]
0.9999     [No]
1.0000     [No]
1.0009     [Yes]
01.0000    [No]
01.0009    [Yes]
001.0000   [No]
011.0000   [Yes]
222.0000   [Yes]
000.9999   [No]

I have been created this regex:

  (?!^1*$)(?!^1*\.0*$)^(?:[1-9]|\d\d\d)(?:\.\d{4,4})$

But it doesn't work for all the cases

regex
3 Answers

You may use this regex to allow up to 3 digits before . and 4 digits after . to match values greater than 1 only:

^(?!0*1\.0+$)(?=0*[1-9])\d{1,3}\.\d{4}$

RegEx Demo

RegEx Details:

  • ^: Start
  • (?!0*1\.0+$): Positive Lookahead condition to assert that we don't have a case of optional zeroes followed by 1.0000
  • (?=0*[1-9]): Positive Lookahead condition to assert that we have at least on non-zero digit ahead
  • \d{1,3}: Match 1 to 3 digits
  • \.: Match a .
  • \d{4}: Match 4 digits
  • $: End
user_5482250

If supported, you might use a negative lookahead to exclude some of the possible matches:

^(?!0*1\.0+$|\d{4}\.)0*[1-9]\d*\.\d{4}$

The pattern matches

  • ^ Start of string
  • (?! Negative lookahead, not directly to the right
    • 0*1\.0+$ Match optional zeroes, a dot and 1 or more zeroes
    • | or
    • \d{4}\. Match 4 digits and a dot
  • ) Close lookahead
  • 0*[1-9]\d* Match optional zeroes, a digit 1-9 and optional digits
  • \.\d{4} match a . and 4 digits
  • $ End of string

Regex demo

user_54249880

Looks like you can use:

^(?!0*\.|0*1\.0*$)\d{1,3}\.\d{4}$

See the demo

  • ^ - Start string anchor.
  • (?! - Open negative lookahead:
    • 0*\. - Prevent 0+ leading zero's upto a literal dot;
    • | - Or:
    • 0*1\.0*$ - Prevent 0+ leading zero's up to a 1 followed by a literal dot and trailing zero's until string ends.
    • ) - Close negative lookahead.
  • \d{1,3}\.\d{4} - 1-3 Leading digits, a literal dot and 4 trailing digits.
  • $ - End string anchor.
user_97581940
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