Regular expression for capturing optionally present groups in Go
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 }
进入
type foo struct { bar foobar baz []foobar bars []foobar }
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`)) }
产生
type foo struct { bar bar baz []foobar bars []foobar }
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`)) }
但这会产生一个缺少所有字段类型的输出:
type Foo struct { Bar Baz Bars }
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`)) }
产生期望的结果
type foo struct { bar foobar baz []foobar bars []foobar }
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