Golang split string by empty newline

Question

I have this JS function that I'm trying to translate to Go:

function splitByEmptyNewline(str) {
  return str
    .replace(/\r\n/g, '\n')
    .split(/^\s*\n/gm);
}

Here's what I got so far:

func splitByEmptyNewline(str string) []string {
    strNormalized := regexp.
        MustCompile("\r\n").
        ReplaceAllString(str, "\n")
    return regexp.
        MustCompile("^s*\n").
        Split(strNormalized, -1)
}

This does not return the same result as the JavaScript version. So I'm wondering what I've missed?

I've tried using both double-quotes " and backward single-quotes ` for the regex.

Can you make the JS one into a StackSnippet with sample input output? — Laurel
– Laurel, Commented Apr 30, 2016 at 17:37

Andreas Koch · Accepted Answer · 2016-04-30 17:53:54Z

2

Your seperator RegEx does not match because you split a complete string and the beginning of that string is not white-space. So instead of ^\s*\n you must use \n\s*\n:

func splitByEmptyNewline(str string) []string {
    strNormalized := regexp.
        MustCompile("\r\n").
        ReplaceAllString(str, "\n")

    return regexp.
        MustCompile(`\n\s*\n`).
        Split(strNormalized, -1)

}

Here here is working example: https://play.golang.org/p/be6Mf3-XNP

edited Apr 30, 2016 at 17:53

answered Apr 30, 2016 at 17:46

Andreas Koch

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1 Comment

Xeoncross Over a year ago

Rather than a regular expression, why not simply remove the windows return? strNormalized := strings.Replace(str, "\r", "", -1)

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