Question

我正在尝试解压缩文件夹并获取该文件夹中存在的文件数，并检查文件数= 32，否则发送一封带有当前文件名的电子邮件，并对当前文件执行infacmds，然后等待32个条件满足。

在此脚本中，我要检查文件数以及现有文件的列数，如果列数为-neq 280，则必须发送电子邮件错误。

列代码是这样的，我必须在主代码中引入它：

$columnCount = ( ( Get-Content "C:\Users\xs15169\Desktop\temp\OEC2_CFLOW.txt" | Select-Object -First 1 ) -Split ',' ).Count
echo "Column count is: $columnCount"

我尝试过的以下代码（＃查找以下类型的文件格式.zip，它将解压缩文件并计数）：

#unzip commands will be here
#LogWrite "Starting of process"
Function LogWrite
{
   Param ([string]$logstring)

   Add-content $log -value "$(Get-Date) $logstring"
}

  try { 
   LogWrite "Start unzipping $zipfilename"
   Remove-Item $unzipfolderdelete | Where { ! $_.PSIsContainer }
    #Unzip $zipfilename $unzipfolder

    Set-Alias sz "C:\Program Files\7-Zip\7z.exe"

    sz x $zipfilename -o"$unzipfolder" -y
    LogWrite "$zipfilename unzipped to $unzipfolder successfully"
    $filecount= (dir $unzipfolder | measure).Count
    if ($filecount -eq 32)
    {
        #LogWrite "File Count 32


        #LogWrite "Executed informatica scripts for condition filecount equal to 32"
        #LogWrite "End of if"
    }
    elseif ($filecount -ne 0)
    {

        LogWrite "File Count " + $filecount
        $list = Get-ChildItem $unzipfolder
        $incompletefilenames =""
        ForEach($n in $list){

        $incompletefilenames = $incompletefilenames + $n.Name + "<br>`n"

        }

        LogWrite $incompletefilenames
        LogWrite "Sending email for filecount not equal to 32"

            $Params = @{ 
               email setup
            }
            Send-MailMessage @Params
            LogWrite "Successfully emailed available file names"

        #Execute Informatica scripts


        LogWrite "Executed informatica scripts for condition filecount not equal to 32"
         LogWrite "End of elseif"
    }
    else
    {
         LogWrite "File Count " + $filecount
             $Params = @{ 
                email setup
            }
            Send-MailMessage @Params
          LogWrite "End of else"
    }
}

Answer 1

为此，您必须找出一些路径。

尝试一下：

Set-Alias sz "C:\Program Files\7-Zip\7z.exe"

$zipFile = "C:\temp\10 yard Fight.zip"

# Get Name of file without extension or path
$zipFolderToCreate = (gci $zipFile).BaseName

# Get path of zip
$path = (gci $zipFile).Directory.FullName

sz x -o"$path\*" "$zipFile"  -r

$filecount= (dir "$path\$zipFolderToCreate" | measure).Count

使用Powershell解压缩文件夹后如何计算每个文件的列

1 个答案: