Replace one occurrence with regexp

Question

I want to replace a only one instance with a regex rather than all of them. How would I do this with Go's regexp library?

input: foobar1xfoobar2x
regex: bar(.)x
replacement: baz$1

ReplaceAllString output: foobaz1foobaz2
ReplaceOneString output: foobaz1foobar2x

That depends on the API. What are you using now? This may help us read the docs for you. — Qtax
– Qtax, Commented May 23, 2013 at 0:02
@Qtax, I'm using Go's regexp library Regexp.ReplaceAllString. An equivalent api for ReplaceOneString doesn't seem to exist. — deft_code
– deft_code, Commented May 23, 2013 at 0:07
You could try the following as a workaround (didn't test it thoroughly) — HamZa
– HamZa, Commented May 23, 2013 at 0:11
@HamZaDzCyberDeV, that solves this one case, but I'm looking for a general solution. — deft_code
– deft_code, Commented May 23, 2013 at 0:17
Actually that solution is very generic. It will always replace the first match with your replacement and leave everything after untouched. — Dominik Honnef
– Dominik Honnef, Commented May 23, 2013 at 0:40

perreal · Accepted Answer · 2013-05-23 01:50:55Z

10

In general, if you use lazy match and use anchors for beginning and end, you can have the replace first behavior:

replace `^(.*?)bar(.*)$` with `$1baz$2`.

Example:

package main

import (
    "fmt"
    "regexp"
)

func main() {
    src := "foobar1xfoobar2x"
    pat := regexp.MustCompile("^(.*?)bar(.*)$")
        repl := "${1}baz$2"
    output := pat.ReplaceAllString(src, repl)
    fmt.Println(output)
}

Output

foobaz1xfoobar2x

edited May 23, 2013 at 1:50

answered May 23, 2013 at 1:11

perreal

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2 Comments

perreal Over a year ago

you need to curly the group variables if they make valid identifier names with the following characters, see the updated answer

deft_code Over a year ago

Wouldn't appending a (.*) to the end of the regex along with a suitable substitution var be enough?

Nykakin · Accepted Answer · 2018-09-13 09:30:21Z

6

I had the same problem. The most clean solution I've come up with:

package main

import (
    "fmt"
    "regexp"
    "strings"
)

func main() {
    re, _ := regexp.Compile("[a-z]{3}")
    s := "aaa bbb ccc"

    // Replace all strings
    fmt.Println(re.ReplaceAllString(s, "000"))

    // Replace one string
    found := re.FindString(s)
    if found != "" {
        fmt.Println(strings.Replace(s, found, "000", 1))
    }
}

Running:

$ go run test.go 
000 000 000
000 bbb ccc

answered Sep 13, 2018 at 9:30

Nykakin

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1 Comment

Kevin Malachowski Over a year ago

This does not work correctly when the regexp is complex enough, e.g. matching on word boundaries will cause this to incorrectly replace an earlier part of the string: go.dev/play/p/gWejkYZ8dB6

noonex · Accepted Answer · 2018-02-21 16:30:56Z

3

I cound't use accepted solution because my pattern was very complicated. I ended up with using ReplaceAllStringFunc : https://play.golang.org/p/ihtuIU-WEYG

package main

import (
    "fmt"
    "regexp"
)

var pat = regexp.MustCompile("bar(.)(x)")

func main() {
    src := "foobar1xfoobar2x"
    flag := false
    output := pat.ReplaceAllStringFunc(src, func(a string) string {
        if flag {
            return a
        }
        flag = true
        return pat.ReplaceAllString(a, "baz$1$2")
    })
    fmt.Println(output)
}

edited Feb 21, 2018 at 16:30

answered Feb 21, 2018 at 16:25

noonex

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