Format a float to n decimal places and no trailing zeros

strconv.FormatFloat(10.900, 'f', -1, 64)

This will result in 10.9.

The -1 as the third parameter tells the function to print the fewest digits necessary to accurately represent the float.

See here: https://golang.org/pkg/strconv/#FormatFloat

I used the below func to achive the same:

//return 45.00 with "45" or 45.50 with "45.5"
func betterFormat(num float32) string {
    s := fmt.Sprintf("%.4f", num)
    return strings.TrimRight(strings.TrimRight(s, "0"), ".")
}

Use the below function to format and control the number of digits.

func FormatFloat(num float64, prc int) string {
    var (
        zero, dot = "0", "."

        str = fmt.Sprintf("%."+strconv.Itoa(prc)+"f", num)
    )
    
    return strings.TrimRight(strings.TrimRight(str, zero), dot)
}

Here is an example:

// Converts 18.24100100 to 18.24
fmt.Println(FormatFloat(18.24100100, 2))

Not sure for Sprintf but to make it worked. Just trim right, first 0 then ..

fmt.Println(strings.TrimRight(strings.TrimRight(fmt.Sprintf("%.2f", 100.900), "0"), ".")) // 100.9
fmt.Println(strings.TrimRight(strings.TrimRight(fmt.Sprintf("%.2f", 100.0), "0"), ".")) // 100