In PowerShell I find myself doing this kind of thing over and over again for matches:
在PowerShell中，我发现自己一遍又一遍地为匹配做这种事情：
some-command | select-string '^(//[^#]*)' |
%{some-other-command $_.matches[0].groups[1].value}
So basically - run a command that generates lines of text, and for each line I want to run a command on a regex capture inside the line (if it matches). Seems really simple. The above works, but is there a shorter way to pull out those regex capture groups? Perl had $1 and so on, if I remember right. Posh has to have something similar, right? I've seen "$matches" references on SO but can't figure out what makes that get set.
所以基本上 - 运行一个生成文本行的命令，对于每一行，我想在行内的正则表达式捕获上运行命令（如果它匹配）。看起来很简单。上面的工作，但有没有更短的方法来拉出这些正则表达式捕获组？ Perl有1美元，依此类推，如果我没记错的话。辣妹必须有类似的东西，对吧？我在SO上看过“$ matches”引用，但无法弄清楚是什么让它得到了设置。
I'm very new to PowerShell btw, just started learning.
我对PowerShell btw很新，刚开始学习。
3 个解决方案
#1
14
You can use the -match operator to reformulate your command as:
您可以使用-match运算符将您的命令重新配置为：
some-command | Foreach-Object { if($_ -match '^(//[^#]*)') { some-other-command $($matches[1])}}
#2
7
Named Replacement
命名替换
'foo bar' -replace '(?<First>foo).+', '${First}'
Returns: foo
返回：foo
Unnamed Replacement
未命名的替换
'foo bar' -replace '(foo).+(ar)', '$2 z$2 $1'
Returns: ar zar foo
返回：ar zar foo
#3
4
You could try this:
你可以试试这个：
Get-Content foo.txt | foreach { some-othercommand [regex]::match($_,'^(//[^#]*)').value }
#1
14
You can use the -match operator to reformulate your command as:
您可以使用-match运算符将您的命令重新配置为：
some-command | Foreach-Object { if($_ -match '^(//[^#]*)') { some-other-command $($matches[1])}}
#2
7
Named Replacement
命名替换
'foo bar' -replace '(?<First>foo).+', '${First}'
Returns: foo
返回：foo
Unnamed Replacement
未命名的替换
'foo bar' -replace '(foo).+(ar)', '$2 z$2 $1'
Returns: ar zar foo
返回：ar zar foo
#3
4
You could try this:
你可以试试这个：
Get-Content foo.txt | foreach { some-othercommand [regex]::match($_,'^(//[^#]*)').value }