正则表达式-解析正则表达式时出错：无效的转义序列：\\ K
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
}
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
}
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
I suggest you take two steps:
1) use regexp find all matches: \d[\d ]+\d
2) filter out which contains 8 digits
(I don’t think you can do this by a single regex in golang)
This will do the job (faster: without any regexp need)
package main
import "fmt"
import "unicode"
import "strings"
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) {
// split msg into lines
lines := strings.FieldsFunc(msg, func(c rune) bool {
return unicode.IsControl(c)
})
// filter numbers
fn := func(ln string) (num string) {
for _, c := range []rune(ln) {
if unicode.IsNumber(c) {
num += string(c)
// fmt.Println(num)
} else if !unicode.IsSpace(c) {
return num
}
}
return num
}
for _, line := range lines {
num := fn(line)
if len(num) == 8 { // 8 numbers in line is the kriterium to accept
return num, nil
}
}
return "eee", nil // Note: Change this later; it's only needed to satisfy func calls above
}