Golang 替换任何和所有换行符

icz*_*cza 5

While using regexp usually yields an elegant and compact solution, often it's not the fastest.

strings.Replacer

Replacer replaces a list of strings with replacements. It is safe for concurrent use by multiple goroutines.

strings.NewReplacer()Replacer.Replace()

Here's how it would look like:

const replacement = "<br>\n"

var replacer = strings.NewReplacer(
    "\r\n", replacement,
    "\r", replacement,
    "\n", replacement,
    "\v", replacement,
    "\f", replacement,
    "\u0085", replacement,
    "\u2028", replacement,
    "\u2029", replacement,
)

func replaceReplacer(s string) string {
    return replacer.Replace(s)
}

Here's how the regexp solution from Wiktor's answer looks like:

var re = regexp.MustCompile(`\r\n|[\r\n\v\f\x{0085}\x{2028}\x{2029}]`)

func replaceRegexp(s string) string {
    return re.ReplaceAllString(s, "<br>\n")
}

The implementation is actually quite fast. Here's a simple benchmark comparing it to the above pre-compiled regexp solution:

const input = "1st\nsecond\r\nthird\r4th\u0085fifth\u2028sixth"

func BenchmarkReplacer(b *testing.B) {
    for i := 0; i < b.N; i++ {
        replaceReplacer(input)
    }
}

func BenchmarkRegexp(b *testing.B) {
    for i := 0; i < b.N; i++ {
        replaceRegexp(input)
    }
}

And the benchmark results:

BenchmarkReplacer-4      3000000               495 ns/op
BenchmarkRegexp-4         500000              2787 ns/op

strings.Replacer

stringstringio.Writerstringstrings.ReplacerReplacer.WriteString()io.Writerstring

Golang 替换任何和所有换行符 |