Return a defined range of characters in Go

Question

Let's say we have a converted float to a string:

"24.22334455667"

I want to just return 6 of the digits on the right of the decimal

I can get all digits, after the decimal this way:

re2 := regexp.MustCompile(`[!.]([\d]+)$`)

But I want only the first 6 digits after the decimal but this returns nothing:

re2 := regexp.MustCompile(`[!.]([\d]{1,6})$`)

How can I do this? I could not find an example of using [\d]{1,6}

Thanks

I think, here better not use regex. Show this: http://stackoverflow.com/questions/18390266/how-can-we-truncate-float64-type-to-a-particular-precision-in-golang and then converting. — , Sep 13 '15 at 17:06

Linear · Answer 1 · 2015-09-13T18:18:19.620

Alternatively...

func DecimalPlaces(decimalStr string, places int) string {
    location := strings.Index(decimalStr, ".")
    if location == -1 {
        return ""
    }
    return decimalStr[location+1 : min(location+1+places, len(decimalStr))]
}

Where min is just a simple function to find the minimum of two integers.

Regular expressions seem a bit heavyweight for this sort of simple string manipulation.

Playground

score 3 · Accepted Answer · answered Sep 13 '15 at 17:01

You must remove the end of the line anchor $ since it won't be a line end after exactly 6 digits. For to capture exactly 6 digits, the quantifier must be

re2 := regexp.MustCompile(`[!.](\d{6})`)

Note that, this would also the digits which exists next to !. If you don't want this behaviour, you must remove the ! from the charcater class like

re2 := regexp.MustCompile(`[.](\d{6})`)

or

For to capture digits ranges from 1 to 6,

re2 := regexp.MustCompile(`[!.](\d{1,6})`)

Return a defined range of characters in Go

2 Answers2