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Home Backend Development Golang Regular expression for capturing optionally present groups in Go

Regular expression for capturing optionally present groups in Go

Feb 09, 2024 pm 02:10 PM

用于捕获 Go 中可选存在的组的正则表达式

php小编百草为您介绍一种用于捕获 Go 中可选存在的组的正则表达式。在Go中,正则表达式是一种强大的工具,可以用来进行字符串匹配和提取。当我们需要匹配一个可选存在的组时,可以使用问号 "?" 来标记这个组的可选性。这种正则表达式的写法简洁明了,非常适合处理需要灵活匹配的情况。通过掌握这种正则表达式的用法,我们可以更好地处理字符串匹配问题,并提取所需的信息。

问题内容

我正在尝试编写一个正则表达式,在表示go代码的字符串中将替换类型的名称,例如bar,并使用更新的名称,例如foobar,但仅限于它作为字段类型出现的地方另一个结构体或该类型的数组。例如,我想进行转换

type foo struct {
    bar  bar
    baz  []bar
    bars []bar
}
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进入

type foo struct {
    bar  foobar
    baz  []foobar
    bars []foobar
}
Copy after login
Copy after login

到目前为止,我已经成功使用此 replaceallstring 转换数组字段类型:

package main

import (
    "fmt"
    "regexp"
)

func main() {
    re := regexp.mustcompile(`(\w+)(\s+)\[\]bar`)

    s := `type foo struct {
    bar  bar
    baz  []bar
    bars []bar
}`

    fmt.println(re.replaceallstring(s, `$1$2[]foobar`))
}
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产生

type foo struct {
    bar  bar
    baz  []foobar
    bars []foobar
}
Copy after login

第一个字段的类型缺少 bar 的替换,也名为 bar。我尝试过将 [] 设为可选,如下所示,

package main

import (
    "fmt"
    "regexp"
)

func main() {
    re := regexp.mustcompile(`(\w+)(\s+)(\[\])?bar`)

    s := `type foo struct {
    bar  bar
    baz  []bar
    bars []bar
}`

    fmt.println(re.replaceallstring(s, `$1$2$3foobar`))
}
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但这会产生一个缺少所有字段类型的输出:

type Foo struct {
    Bar
    Baz
    Bars
}
Copy after login

这里出了什么问题? (我可以使用带有两个不同正则表达式的两遍方法,但我更愿意一次性实现这一点)。

解决方法

事实证明,第三个引用需要是 ${3},而不是 $3

package main

import (
    "fmt"
    "regexp"
)

func main() {
    re := regexp.mustcompile(`(\w+)(\s+)(\[\])?bar`)

    s := `type foo struct {
    bar  bar
    baz  []bar
    bars []bar
}`

    fmt.println(re.replaceallstring(s, `$1$2${3}foobar`))
}
Copy after login

产生期望的结果

type foo struct {
    bar  foobar
    baz  []foobar
    bars []foobar
}
Copy after login
Copy after login

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