Remove insignificant trailing zeros from a number?

I first used a combination of matti-lyra and gary's answers:

r=(+n).toFixed(4).replace(/\.0+$/,'')

Results:

• 1234870.98762341: "1234870.9876"
• 1230009100: "1230009100"
• 0.0012234: "0.0012"
• 0.1200234: "0.12"
• 0.000001231: "0"
• 0.10001: "0.1000"
• "asdf": "NaN" (so no runtime error)

The somewhat problematic case is 0.10001. I ended up using this longer version:

    r = (+n).toFixed(4);
    if (r.match(/\./)) {
      r = r.replace(/\.?0+$/, '');
    }

• 1234870.98762341: "1234870.9876"
• 1230009100: "1230009100"
• 0.0012234: "0.0012"
• 0.1200234: "0.12"
• 0.000001231: "0"
• 0.10001: "0.1"
• "asdf": "NaN" (so no runtime error)

Update: And this is Gary's newer version (see comments):

r=(+n).toFixed(4).replace(/([0-9]+(\.[0-9]+[1-9])?)(\.?0+$)/,'$1')

This gives the same results as above.

If you convert it to a string it will not display any trailing zeros, which aren't stored in the variable in the first place since it was created as a Number, not a String.

var n = 1.245000
var noZeroes = n.toString() // "1.245" 

I had a similar instance where I wanted to use .toFixed() where necessary, but I didn't want the padding when it wasn't. So I ended up using parseFloat in conjunction with toFixed.

toFixed without padding

parseFloat(n.toFixed(4));

Another option that does almost the same thing
This answer may help your decision

Number(n.toFixed(4));

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toFixed will round/pad the number to a specific length, but also convert it to a string. Converting that back to a numeric type will not only make the number safer to use arithmetically, but also automatically drop any trailing 0's. For example:

var n = "1.234000";
    n = parseFloat(n);
 // n is 1.234 and in number form

Because even if you define a number with trailing zeros they're dropped.

var n = 1.23000;
 // n == 1.23;