我正在尝试编译一个正则表达式,以便可以使用Go从字符串中提取8位数字,其中数字之间有/无空格。由于某些原因,编译失败。我应该用什么代替K?
validAcc, err := regexp.Compile(`[ ]\K(?<!\d )(?=(?: ?\d){8})(?!(?: ?\d){9})\d[ \d]+\d`)
if err != nil {
return
}
带有示例数据的更多代码
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
}
考虑到go regexp r2不支持任何向前/向后搜索,您可以先尝试使用一个更简单的表达式吗:
c, err := regexp.Compile(`\b\d{8}\b`)
以您的情况(操场),这将工作
(\d\d ){4}
validAcc, err := regexp.Compile(`(\d\d ){4}`)
要么:
(\d\d ?){4} # matches '33 1133 06 Oth'
validAcc, err := regexp.Compile(`(\d\d ?){4}`)
再次,我先尝试一个简单的正则表达式,然后再尝试更复杂的选项:它将取决于您必须解析的数据。
对于更复杂的情况,单独的正则表达式可以帮助您捕获组中的数据,然后您需要提取找到的数字(这意味着您需要将后处理添加到正则表达式中):
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)
参观游乐场
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