windows - 如何消除脚本 block 中的重复代码？

Question

Powershell 新手。我正在编写一个脚本来监视目录中的文件并向控制台报告更改。

我注意到我用于 FS“观察者”的脚本 block 中有相当多的代码重复。

这是脚本 block 的片段。如果需要，我可以发布整个脚本，它有点长，而且还有更多内容。

# Filter all files
$filter = "*.*"
$watcher = New-Object IO.FileSystemWatcher $watchdir, $filter -Property @{
    IncludeSubdirectories = $true
    EnableRaisingEvents = $true
}

# File creation
Register-ObjectEvent $watcher Created -SourceIdentifier Created -Action {
    $path = $Event.SourceEventArgs.FullPath
    $name = $Event.SourceEventArgs.Name
    $changeType = $Event.SourceEventArgs.ChangeType
    $timeStamp = $Event.TimeGenerated
    $message = "The file '$name' was '$changeType' at '$timeStamp'"
    Write-Host $message
}

# File change
Register-ObjectEvent $watcher Changed -SourceIdentifier Changed -Action {
    $path = $Event.SourceEventArgs.FullPath
    $name = $Event.SourceEventArgs.Name
    $changeType = $Event.SourceEventArgs.ChangeType
    $timeStamp = $Event.TimeGenerated
    $message = "The file '$name' was '$changeType' at '$timeStamp'"
    Write-Host $message
}

# File rename
...

# File delete
...

是否有一个好的模式或更好的编写方法来减少代码量？

Answer 1

在您的示例中，脚本 block 完全相同，因此在这种情况下，只需将它们放入变量中并传递即可:

$action = {
    $path = $Event.SourceEventArgs.FullPath
    $name = $Event.SourceEventArgs.Name
    $changeType = $Event.SourceEventArgs.ChangeType
    $timeStamp = $Event.TimeGenerated
    $console_message = "The file '$name' was '$changeType' at '$timeStamp'"
    Write-Host message
}

# File creation
Register-ObjectEvent $watcher Created -SourceIdentifier Created -Action $action

# File change
Register-ObjectEvent $watcher Changed -SourceIdentifier Changed -Action $action

如果您发布的示例中脚本 block 存在冗余但它们并不完全相同，那么展示如何最好地抽象它的示例会更容易。可能没有一种放之四海而皆准的解决方案。