正则表达式-解析正则表达式时出错：无效的转义序列：\\ K

Question

I'm trying to compile a regex so that I can extract an 8 digit number with/without spaces between the digits from a string using Go. For some reasons the compilations fails. What should I repalce K with ?

validAcc, err := regexp.Compile(`[ ]\K(?<!\d )(?=(?: ?\d){8})(?!(?: ?\d){9})\d[ \d]+\d`)
if err != nil {
    return
}

Play it here

More code with sample data

package main

import "strings"
import "regexp"
import "fmt"

func main() {

    msg := ` 12 34 56 78 //the number we need
 12 3455678 90123455 // the number we don't need`

    acc, err := accFromText(msg)
    if err != nil {
        panic(err)
    }
    exAcc := "12345678"
    if acc != exAcc {
        fmt.Printf("expected %v, received %v", exAcc, acc)
    }

    msg = `
More details here
1234567 12345 123456789 asd
12000000000 a number we don't need 
 12 3456 78 //this is the kind of number we need
 12 3455678 90123455 // the number we don't need`

    acc, err = accFromText(msg)
    if err != nil {
        panic(err)
    }
    exAcc = "12345678"
    if acc != exAcc {
        fmt.Printf("expected %v, received %v", exAcc, acc)
    }

}

func accFromText(msg string) (accNumber string, err error) {
    validAcc, err := regexp.Compile(`[ ]\K(?<!\d )(?=(?: ?\d){8})(?!(?: ?\d){9})\d[ \d]+\d`)
    if err != nil {
        return
    }
    accNumber = string(validAcc.Find([]byte(msg)))
    accNumber = strings.Replace(accNumber, " ", "", -1)
    return
}

Answer 1

Considering the go regexp r2 doesn't support any lookbehind/ahead, could you try a simpler expression first:

c, err := regexp.Compile(`\b\d{8}\b`)

In your case (playground), this would work

(\d\d ){4}
validAcc, err := regexp.Compile(`(\d\d ){4}`)

Or:

(\d\d ?){4} # matches '33 1133 06 Oth'
validAcc, err := regexp.Compile(`(\d\d ?){4}`)

Again, I try first a simple regexp, before trying more complex option: it will depend on the data you have to parse.

---

For a more complex case, the regexp alone can help you capture the data in a group, and then you need to extract the number found (meaning you ned to add post-processing to your regexp):

validAcc, err := regexp.Compile(`[^\d]((\d\d ?){4})[^\d]`)
if err != nil {
    return
}
accNumber = string(validAcc.Find([]byte(msg)))[1:]
accNumber = accNumber[:len(accNumber)-1]
accNumber = strings.Replace(accNumber, " ", "", -1)

See playground