How to simulate negative lookbehind in Go

Question

I\'m trying to write a regex that can extract a command, here\'s what I\'ve got so far using a negative lookbehind assertion:

Answer 1

You can actually match the preceding character (or the beginning of line) and use a group to get the desired text in a subexpression.

Regex

(?:^|[^@#/])\b(\w+)

• (?:^|[^@#/]) Matches either ^ the beginning of line or [^@#/] any character except @#/
• \b A word boundary to assert the beginning of a word
• (\w+) Generates a subexpression
  • and matches \w+ any number of word characters

Code

cmds := []string{
    `/msg @nickname #channel foo bar baz`,
    `#channel @nickname foo bar baz /foo`,
    `foo bar baz @nickname #channel`,
    `foo bar baz#channel`}

regex := regexp.MustCompile(`(?:^|[^@#/])\b(\w+)`)


// Loop all cmds
for _, cmd := range cmds{
    // Find all matches and subexpressions
    matches := regex.FindAllStringSubmatch(cmd, -1)

    fmt.Printf("`%v` \t==>\n", cmd)

    // Loop all matches
    for n, match := range matches {
        // match[1] holds the text matched by the first subexpression (1st set of parentheses)
        fmt.Printf("\t%v. `%v`\n", n, match[1])
    }
}

Output

`/msg @nickname #channel foo bar baz`   ==>
    0. `foo`
    1. `bar`
    2. `baz`
`#channel @nickname foo bar baz /foo`   ==>
    0. `foo`
    1. `bar`
    2. `baz`
`foo bar baz @nickname #channel`    ==>
    0. `foo`
    1. `bar`
    2. `baz`
`foo bar baz#channel`   ==>
    0. `foo`
    1. `bar`
    2. `baz`

Playground
http://play.golang.org/p/AaX9Cg-7Vx