Golang Split String with Regexp-CodePudding

I have a string and I split the string with regexp package. However I could not create a regexp I need.

The regexp code lines:

v = "@636e0e0eac0bd25cd5df92a3$636e0e49ac0bd25cd5df92a5.result.result[0].code.xTrim()"
r, _ := regexp.Compile(`(@[a-zA-Z0-9] \$[a-zA-Z0-9] )(.*)`)
variableParts := r.FindStringSubmatch(v)

VariablesParts array:

0: "@636e0e0eac0bd25cd5df92a3$636e0e49ac0bd25cd5df92a5.result.result[0].code.xTrim()"

1: "@636e0e0eac0bd25cd5df92a3$636e0e49ac0bd25cd5df92a5"

2: ".result.result[0].code.xTrim()"

This is my desired result:

Array length can change but array should include this 4 strings.

0: "@636e0e0eac0bd25cd5df92a3$636e0e49ac0bd25cd5df92a5.result.result[0].code.xTrim()"

1: "@636e0e0eac0bd25cd5df92a3$636e0e49ac0bd25cd5df92a5"

2: ".result.result[0].code.xTrim()"

3: ".xTrim()"

How Can I get this result. Which regexp string I need?

UPDATE for new desired result:

0: "@636e0e0eac0bd25cd5df92a3$636e0e49ac0bd25cd5df92a5.result.result[0].code.xTrim()"

1: "@636e0e0eac0bd25cd5df92a3$636e0e49ac0bd25cd5df92a5"

2: "result"

3: "result[0]"

4: "code"

5: "xTrim()"

CodePudding user response：

You can replace the last (.*) part with (.*(\.[^.]*))$ and use

(@[a-zA-Z0-9] \$[a-zA-Z0-9] )(.*(\.[^.]*))$

See the regex demo. Details:

(@[a-zA-Z0-9] \$[a-zA-Z0-9] ) - Group 1: a @ char, then one or more alphanumeric chars, then a # char and then again one or more alphanumeric chars
(.*(\.[^.]*)) - Group 2: any zero or more chars other than line break chars as many as possible, and then Group 3: a ., zero or more chars other than a . char
$ - at the end of string. Remove if there can be more text on the right.

CodePudding user response：

A better option would be to just use strings.Split:

package main

import (
   "fmt"
   "strings"
)

func main() {
   s := "@636e0e0eac0bd25cd5df92a3$636e0e49ac0bd25cd5df92a5.result.result[0].code.xTrim()"
   a := strings.Split(s, ".")
   fmt.Println(a)
}

https://godocs.io/strings#Split