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Here's the scenario: I'm getting .9999999999999999 when I should be getting 1.0.
I can afford to lose a decimal place of precision, so I'm using .toFixed(15), which kind of works.
The rounding works, but the problem is that I'm given 1.000000000000000.
Is there a way to round to a number of decimal places, but strip extra whitespace?
Note: .toPrecision isn't what I want; I only want to specify how many numbers after the decimal point.
Note 2: I can't just use .toPrecision(1) because I need to keep the high precision for numbers that actually have data after the decimal point. Ideally, there would be exactly as many decimal places as necessary (up to 15).
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To remove the trailing zeros from a number, pass the number to the parseFloat() function. The parseFloat function parses the provided value, returning a floating point number, which automatically removes any trailing zeros.
To determine the number of significant figures in a number use the following 3 rules: Non-zero digits are always significant. Any zeros between two significant digits are significant. A final zero or trailing zeros in the decimal portion ONLY are significant.
>>> parseFloat(0.9999999.toFixed(4)); 1 >>> parseFloat(0.0009999999.toFixed(4)); 0.001 >>> parseFloat(0.0000009999999.toFixed(4)); 0 Yes, there is a way. Use parseFloat().
parseFloat((1.005).toFixed(15)) //==> 1.005 parseFloat((1.000000000).toFixed(15)) //==> 1 See a live example here: http://jsfiddle.net/nayish/7JBJw/
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