Question

更改文件A时，我需要打开文件B，然后让文件B打开文件A，而无需再次触发监视程序。我的想法是，可以使用'$ changeAction'中的if语句来执行此操作，该语句检查是否首先打开了文件B。

我想这就像：

如果（文件B已打开，则什么都不做，否则打开文件B）

如何在Powershell中编写此代码？

 Function Register-Watcher {
        param ($folder)
        $filter = "*.xlsx"
        $folder = "\\powershell\watcher\test\folder"

        $watcher = New-Object IO.FileSystemWatcher $folder, $filter -Property @{ 
            IncludeSubdirectories = $false
            EnableRaisingEvents = $true


        }

        $changeAction = [scriptblock]::Create('

            $path = $Event.SourceEventArgs.FullPath
            $name = $Event.SourceEventArgs.Name
            $changeType = $Event.SourceEventArgs.ChangeType
            $timeStamp = $Event.TimeGenerated

            Write-Host "The file $name was $changeType at $timeStamp"

            $Excel = New-Object -ComObject Excel.Application
            $Excel.Workbooks.Open("\\powershell\watcher\test\folder\fileB.xlsm") #this should be the 'else' in the 'if File B is open#

        ')

        Register-ObjectEvent $Watcher "Changed" -Action $changeAction
    }

     Register-Watcher "\\powershell\watcher\test\folder\fileA.xlsx"
     $Change

Answer 1

为此，您可以检查是否正在运行任何打开了特定文件的Excel进程。

Get-CimInstance Win32_Process -Filter "CommandLine like '%filepath.xlsx%'"

根据上述表达式的输出，您可以决定是否打开文件。

如果已打开Excel文件，如何防止Powershell打开它？

1 个答案: