CODEWITHSUNDEEP
javascriptregex
jjkl
jjkl

Reputation: 403

Match using first part of regex if decimal exists and second part if none

I'm trying to match the leading zeros of a string to replace with an empty string. I can have a scenario with leading zeros such that there is a decimal: 0000000.005 or without: 000000000005.

I've come up with: /.+?(?=0\.)|.+?(?=[0-9]+)/.

The first part matches all the leading zeroes up until I reach a 0. pattern (i.e. 0000000.005 becomes 0.005). The second part applies the same lazy, positive lookahead approach to leading zeros without a decimal (i.e. 000000000005 becomes 5).

Since the second part (.+?(?=[0-9]+)) is stronger than the first, it will always turn 0000000.005 into 5. Is there a way to, in one expression, only match using the first section if there is, say the presence of a decimal, and to use the other expression if not?

Thanks!

Upvotes: 1

Views: 56

Answers (2)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626747

If you need a regex solution, you may use

s.replace(/^0+(?=0\.|[1-9])/, '')

See the regex demo.

Details

  • ^ - start of string
  • 0+ - 1+ zeros
  • (?=0\.|[1-9]) - the next chars should be 0. or a digit from 1 to 9.

JS demo:

var strs = ['000000000005', '0000000.005'];
for (var s of strs) {
  console.log(s, '=>', s.replace(/^0+(?=0\.|[1-9])/, ''));
  console.log(s, '=>', Number(s)); // Non-regex way
}

Upvotes: 2

Pushprajsinh Chudasama
Pushprajsinh Chudasama

Reputation: 7949

You can use the replace() fucntion in regex.
Here is the demo code . 

var str = "00000000.00005";
var str1 = "0000000005";
if(str.includes(".")) //case 1
{
	str = str.replace(/0+\./g, '0.');
console.log("str ==>", str);
}
if(!str1.includes("."))//case 2 . 
{ 
		str1 = str1.replace(/0+\.|0+/g, '');
console.log("str 1 ===>", str1);

}

Upvotes: 0

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