Golang regexpp：获取带有右括号的单词

发布于04月25日

需要从字符串中提取单词的帮助.

STRING	WORD NEEDS TO BE FETCHED
"A = 2&& B = 2 && func1(A) && func2(B)"	func1, func2
"func1(A)"	func1
"B = 2 && func2(A)"	func2

我try 过遵循regex

package main

import (
    "fmt"
    "regexp"
)

func main() {
    var re = regexp.MustCompile(`\s+(?:\w+_\w+)`)
    var str = `func1(A)`

    for i, match := range re.FindAllString(str, -1) {
        fmt.Println(match, "found at index", i)
    }
}

但如果单词eg:对于字符串"func1(A)"来说，之前没有空间，则该regex提供空结果.

当单词前面有一个空位时，这会按预期工作，例如对于字符串"B = 2 func2(A)"，它会按预期给出func2.

如果有任何帮助，我们将不胜感激

推荐答案

要查找所有请求的单词，您可以使用以下代码:

package main

import (
    "fmt"
    "regexp"
)

func main() {
    var re = regexp.MustCompile(`(^|\s)(\w+)\(.*?\)`)
    var str = `A = 2&& B = 2 && func1(A) && func2(B)
    func1(A)
    B = 2 && func2(A) func()`

    for _, match := range re.FindAllStringSubmatch(str, -1) {
        fmt.Println(match[2], "found")
    }
}

regex的解释:

(^|\s) - The beginning of a line or a whitespace.
(\w+)  - Minimum 1 word character, with as many as possible.
\(     - The opening bracket literally.
.*?    - Any character, zero or few occurencies as possible.
\)     - The closing bracket literally.

您可以测试代码here