如何编写正则表达式来压缩文件夹路径名称更改？

Question

我正在尝试为 JavaScript 编写正则表达式。

我有文件夹的文件名，我需要从 git 更改中合并它们。 例如 输入：

"9  1   {Folder_Old => Folder}/FileTest1.cs" 
"0  9   File{a => t}est2.cs" 
"-  -   F{a => i}leT{b => e}st.d{c => l}l" 
"9  1   {test/File.cs => test/File1.cs}" 

每行的预期输出：

"9  1   Folder/FileTest1.cs" 
"0  9   Filetest2.cs" 
"-  -   FileTest.dll" 
"9  1   test/File1.cs" 

这是我迄今为止尝试过的：

var myentry = '9 9 {Folder_Old => Folder}/FileTest1.cs'

//find { A => B }

//replace { A => B } with B

let result = myentry.match(/{.+=>.+}/g);
console.log(result[0]) //"{Folder_Old => Folder}"

result[0] = result[0].replace('{', '')
result[0] = result[0].replace('}', '')
console.log(result[0]) //"Folder_Old => Folder"

var me = result[0].split(' ')
console.log(me) //["Folder_Old", "=>", "Folder"]

var he = myentry.replace(/{.+=>.+}/g, me[2])
console.log(he) //"9 9 Folder/FileTest1.cs"

如何更改我的算法以涵盖所有情况？

Answer 1

如果您匹配替换项（在{} 内），捕获新值，您可以将匹配项替换为捕获的组。

匹配 {.*? => ([^}]+?)} 并替换为 $1。

来自 regex101 的解释：

{ matches the character { literally (case insensitive)
. matches any character (except for line terminators)
*? matches the previous token between zero and unlimited times, as few times as possible, expanding as needed (lazy)
 =>  matches the characters  =>  literally (case insensitive)
1st Capturing Group ([^}])
Match a single character not present in the list below [^}]
+? matches the previous token between one and unlimited times, as few times as possible, expanding as needed (lazy)
} matches the character } literally (case insensitive)
} matches the character } literally (case insensitive)

var samples = [
      '"9  1   {Folder_Old => Folder}/FileTest1.cs"',
      '"0  9   File{a => t}est2.cs"',
      '"-  -   F{a => i}leT{b => e}st.d{c => l}l"',
      '"9  1   {test/File.cs => test/File1.cs}"'
    ];

samples.forEach(s => console.log(s.replace(/{.*? => ([^}]+?)}/gmi, '$1')));

See it at regex101.