在PowerShell中

Question

取下7zip的输出流中的空行，我提取.tgz的档案与PowerShell和7zip的。 7zip有一个强制详细模式并污染我的屏幕。我只想要显示错误消息。

代码：

Write-Host "Extracting "$strArgument 

    # extract tgz to tar 
    & "$7z" x "-y" $strArgument | FIND /V "ing " | FIND /V "Igor Pavlov" | FIND /V "Processing " | FIND /V "Everything is Ok" | FIND /V "Folders: " | FIND /V "Files: " | FIND /V "Size: " | FIND /V "Compressed: " | FIND /V "Processing archive: " 

    # path to .tar 
    $strArgument = $strArgument.Replace(".tgz", ".tar") 
    $strArgument = $strArgument.Replace("i\data\", "") 

    # extract tar to file 
    #Write-Host $strArgument 
    & "$7z" x "-y" $strArgument | FIND /V "ing " | FIND /V "Igor Pavlov" | FIND /V "Processing " | FIND /V "Everything is Ok" | FIND /V "Folders: " | FIND /V "Files: " | FIND /V "Size: " | FIND /V "Compressed: " | FIND /V "Processing archive: " 

这让我有很多的空行输出：

Extracting Q:\mles\etl-i_test\i\data\divider-bin.tgz 










Extracting Q:\mles\etl-i_test\i\data\divider-conf.tgz 
... 

不过，我只是想：

Extracting Q:\mles\etl-i_test\i\data\divider-bin.tgz 
Extracting Q:\mles\etl-i_test\i\data\divider-conf.tgz 

我已经与FIND除去标准的消息

如何从我的流中删除空行？是否有强大的FIND开关？

Answer 1

find没有一个这样的开关，但你可以使用findstr用正则表达式代替：

... | findstr /r /v "^$" 

Answer 2

我知道

最简单的方法就是通过-match运行结果，并过滤掉任何不包含非空白：

Write-Host "Extracting "$strArgument 

    # extract tgz to tar 
    & "$7z" x "-y" $strArgument | FIND /V "ing " | FIND /V "Igor Pavlov" | FIND /V "Processing " | FIND /V "Everything is Ok" | FIND /V "Folders: " | FIND /V "Files: " | FIND /V "Size: " | FIND /V "Compressed: " | FIND /V "Processing archive: " 

    # path to .tar 
    $strArgument = $strArgument.Replace(".tgz", ".tar") 
    $strArgument = $strArgument.Replace("i\data\", "") 

    # extract tar to file 
    #Write-Host $strArgument 
    (& "$7z" x "-y" $strArgument | FIND /V "ing " | FIND /V "Igor Pavlov" | FIND /V "Processing " | ND /V "Everything is Ok" | FIND /V "Folders: " | FIND /V "Files: " | FIND /V "Size: " | FIND /V "Compressed: " | FIND /V "Processing archive: ") -match '\S'