Remove insignificant trailing zeros from a number?
Have I missed a standard API call that removes trailing insignificant zeros from a number?
Ex.
var x = 1.234000 // to become 1.234;
var y = 1.234001; // stays 1.234001
Number.toFixed() and Number.toPrecision() are not quite what I'm looking for.
如果将其转换为字符串，它将不会显示任何尾随零，这些尾随零并不存储在变量中，因为它是作为数字创建的，而不是字符串。
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));
toFixed will truncate/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;
I first used a combination of matti-lyra and gary's answers:
r=(+n).toFixed(4).replace(/.0+$/,'')
Results:
The somewhat problematic case is 0.10001. I ended up using this longer version:
r = (+n).toFixed(4);
if (r.match(/./)) {
r = r.replace(/.?0+$/, '');
}
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.
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