2015-11-22 06:03
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按照要求的格式将float转换为go lang中的字符串。

I have 4 float values(startLat,startLon,endLat,endLon)in go . I want to append(replace) these values to below string:

var etaString = []byte(`{"start_latitude":"` + startLat + `","start_longitude":"` + startLon + `","end_latitude":"` + endLat + `","end_longitude":"` + endLon }`)

I have to typecast them to string before doing so.

startLat := strconv.FormatFloat(o.Coordinate.Longitude, 'g', 1, 64)

However when I do so, I get the values of these parameters as "4e+01 -1e+02 4e+01 -1e+02" But I just want something like this : "64.2345" .

How can I achieve this ? TIA :)

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我在go中有4个浮点值(startLat,startLon,endLat,endLon)。 我想将这些值附加(替换)到以下字符串:

  var etaString = [] byte(`{“ start_latitude”:“`+ startLat +`”,“ start_longitude  “:”`+ startLon +`“,” end_latitude“:”`+ endLat +`“,” end_longitude“:”`+ endLon}`)

  startLat:= strconv.FormatFloat(o.Coordinate.Longitude,'g',1,64)

但是,当我这样做时,我得到的这些参数的值为“ 4e + 01 -1e + 02 4e + 01 -1e + 02” 但是我只想要这样的东西:“ 64.2345”。

如何实现? TIA :)

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1条回答 默认 最新

  • dtqscvrzn968518698 2015-11-22 06:11

    Package strconv

    import "strconv" > func FormatFloat

    func FormatFloat(f float64, fmt byte, prec, bitSize int) string

    FormatFloat converts the floating-point number f to a string, according to the format fmt and precision prec. It rounds the result assuming that the original was obtained from a floating-point value of bitSize bits (32 for float32, 64 for float64).

    The format fmt is one of 'b' (-ddddp±ddd, a binary exponent), 'e' (-d.dddde±dd, a decimal exponent), 'E' (-d.ddddE±dd, a decimal exponent), 'f' (-ddd.dddd, no exponent), 'g' ('e' for large exponents, 'f' otherwise), or 'G' ('E' for large exponents, 'f' otherwise).

    The precision prec controls the number of digits (excluding the exponent) printed by the 'e', 'E', 'f', 'g', and 'G' formats. For 'e', 'E', and 'f' it is the number of digits after the decimal point. For 'g' and 'G' it is the total number of digits. The special precision -1 uses the smallest number of digits necessary such that ParseFloat will return f exactly.

    Use a precision of -1, not 1. Use a format of f, not g to avoid exponent form for large exponents (see HectorJ's comment).

    startLat := strconv.FormatFloat(o.Coordinate.Longitude, 'f', -1, 64)

    For example,

    package main
    import (
    func main() {
        f := 64.2345
        s := strconv.FormatFloat(f, 'g', 1, 64)
        s = strconv.FormatFloat(f, 'f', -1, 64)


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