如何使用正则表达式将所有内容放在golang的括号内 [英] How to get everything within brackets in golang with regex

问题描述

我正在尝试在golang正则表达式中的以下sql语句的括号中获取所有内容.

I am trying to get everything within the outside brackets of the following sql statement in golang regular expressions.

Categories
(// = outside bracket
PersonID int,
LastName varchar(255),
FirstName varchar(255),
Address varchar(255),
City varchar(255)
)//=outside bracket

我将如何使用正则表达式仅识别外括号并返回外括号之间的所有内容?

how would I use regex to only identify the outer brackets and return everything in between the outside brackets?

推荐答案

您所需要做的就是找到第一个(，然后将所有字符匹配到最后一个)与

All you need is to find the first (, then match any characters up to the last ) with

`(?s)\((.*)\)`

详细信息:

• (?s)-允许.匹配任何字符，包括换行符
• \(-文字( char
• (.*)-子匹配1捕获任何零个或多个字符
• \)-文字)符号.

• (?s) - allow . to match any character, including a newline
• \( - a literal ( char
• (.*) - Submatch 1 capturing any zero or more characters
• \) - a literal ) symbol.

请参见开始演示:

package main

import (
    "fmt"
    "regexp"
)

func main() {
    s := `Categories
(// = outside bracket
PersonID int,
LastName varchar(255),
FirstName varchar(255),
Address varchar(255),
City varchar(255)
)//=outside bracket`
    re := regexp.MustCompile(`(?s)\((.*)\)`)

    m := re.FindAllStringSubmatch(s,-1)
    fmt.Printf("Capture value: %s", m[0][1])
}

这篇关于如何使用正则表达式将所有内容放在golang的括号内的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！