从输出中删除路径 [英] Remove path from output

问题描述

在PowerShell中使用以下Select-String命令:

Using the following Select-String command in PowerShell:

Select-String -Path E:\Documents\combined0.txt -Pattern "GET /ccsetup\.exe" -AllMatches > E:\Documents\combined3.txt

创建一个输出文件，每行以路径和文件名开头，后跟冒号.例如:

creates an output file with each line starting with the path and file name followed by a colon. For example:

E:\Documents\combined0.txt:255:255.255.255 - - [31/Dec/2014:04:15:16 -0800] "GET /ccsetup.exe HTTP/1.1" 301 451 "-" "Mozilla/5.0 (compatible; SearchmetricsBot; http://www.xxxx.com/en/xxxx-bot/)"

如何删除结果中的输出文件路径名，输出文件名和冒号?

How do I get rid of the output file path name, output file name and colon in the results?

推荐答案

Select-String输出一个对象，您可以从中选择所需的属性. Get-Member命令将向您显示这些对象成员，如果您通过管道将其插入，例如:

Select-String outputs an object from which you can pick off properties that you want. The Get-Member command will show you these object members if you pipe into it e.g.:

Select-String -Path E:\Documents\combined0.txt -Pattern "GET /ccsetup\.exe" -AllMatches  | 
    Get-Member

其中一个属性是Line.因此，请尝试这种方式:

One of those properties is Line. So try it this way:

Select-String -Path E:\Documents\combined0.txt -Pattern "GET /ccsetup\.exe" -AllMatches | 
    Foreach {$_.Line} > E:\Documents\combined3.txt

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