javascript - Regular expression to enforce 2 digits after decimal point - Stack Overflow

admin管理员组

文章数量:1186450

I need to validate a numeric string with JavaScript, to ensure the number has exactly two decimal places.

The validation will pass only if

1. the number has precisely two decimal places
2. there is at least one digit before the decimal point. (could be zero)
3. the number before the decimal point can not begin with more than one zero.

Valid numbers:

0.01
0.12
111.23
1234.56
012345.67
123.00
0.00

Invalid numbers:

.12
1.1
0.0
00.00
1234.
1234.567
1234
00123.45
abcd.12
12a4.56
1234.5A

I have tried the regular expression [0-9][\.][0-9][0-9]$, but it allows letters before decimal point like 12a4.56.

I need to validate a numeric string with JavaScript, to ensure the number has exactly two decimal places.

The validation will pass only if

1. the number has precisely two decimal places
2. there is at least one digit before the decimal point. (could be zero)
3. the number before the decimal point can not begin with more than one zero.

Valid numbers:

0.01
0.12
111.23
1234.56
012345.67
123.00
0.00

Invalid numbers:

.12
1.1
0.0
00.00
1234.
1234.567
1234
00123.45
abcd.12
12a4.56
1234.5A

I have tried the regular expression [0-9][\.][0-9][0-9]$, but it allows letters before decimal point like 12a4.56.

• javascript
• regex

Share Improve this question Follow edited Nov 3, 2012 at 12:14 phant0m 16.9k66 gold badges5151 silver badges8484 bronze badges asked Nov 3, 2012 at 10:39 Manas SahaManas Saha 1,49799 gold badges2929 silver badges4545 bronze badges

Add a comment  | 

7 Answers 7

Sorted by: Reset to default 8

. matches any character, it does not do what you think it does. You have to escape it. Also, you have two more errors; try

^[0-9]+\.[0-9][0-9]$

instead, or even better, use \d for decimal digits:

^\d+\.\d\d$

This covers all requirements:

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

• the number has precisely two decimal places
  • Trivially satisfied due to the non-optional \.\d\d$

The other two conditions can be restated as follows:

• The number before the decimal points is either a zero
• or a number with exactly one zero, then a number that does not start with zero

This is covered in these two cases:

• 0
• 0?[1-9]\d*

You don't need regular expressions for this.

JavaScript has a function toFixed() that will do what you need.

var fixedtotwodecimals = floatvalue.toFixed(2);

i used this

^[1-9][1-9]*[.]?[1-9]{0,2}$

• 0 not accept

• 123.12 accept but 123.123 not accept

• 1 accept

• 12213123 accept

• sdfsf not accept

• 15.12 accept

• 15@12 not accept

• 15&12 not accept

var values='0.12';

document.write(values.match(/\d+[.]+\d+\d/));

change value as you want and check it

Here it is:

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

Try This Code

pattern="[0-9]*(\.?[0-9]{1,2}$)?"

• 1 Valid

• 1.1 Valid

• 1.12 Valid

• 1.123 not Valid

• only number Valid

  pattern="[0-9]*(.?[0-9]{2}$)?"

• 1 Valid

• 1.1 not Valid

• 1.12 Valid

• 1.123 not Valid

• only number Valid

本文标签： javascriptRegular expression to enforce 2 digits after decimal pointStack Overflow