Question

我正在编写一个删除旧sql备份的脚本，但只有在有新版本的情况下才会删除。我写了下面的脚本，但不确定我是否应该以这种方式处理$d[0] -eq $null部分。我尝试进行计数，但逻辑仍然失败，因为如果$dir | Get-ChildItem | Where-Object {$_.LastWriteTime -gt (Get-Date).AddDays($days) -and $_.extension.ToString() -eq ".bak"}没有返回$d仍然没有计数1.那么，还有另一种方式或写这个吗？

Set-Location \\Server\Backups

$days = -2
$list = @()

$DBFolders = Get-ChildItem | Where-Object {$_.PSIsContainer -eq $true }

Foreach ($dir in $DBFolders)
{
    $d = @()
    $d += $dir | Get-ChildItem | Where-Object {$_.LastWriteTime -gt (Get-Date).AddDays($days) -and $_.extension.ToString() -eq ".bak"}

    if ($d[0] -ne $null)
        {
            $list += $dir | Get-ChildItem | Where-Object {$_.LastWriteTime -lt (Get-Date).AddDays($days) -and $_.extension.ToString() -eq ".bak" -or $_.extension.tostring() -eq ".trn"}
        }
}

$list | remote-item

Answer 1

只需测试$d是否评估为$true或$false - 空集合始终评估为$false：

$d = @()
$d += $dir | Get-ChildItem | Where-Object {$_.LastWriteTime -gt (Get-Date).AddDays($days) -and $_.extension.ToString() -eq ".bak"}

if($d){
    # $d is not empty
}

Answer 2

您可以尝试以下操作：

$ts = (Get-Date).AddDays(-2); # so we don't call it dozens of times
$dirsToClean = Get-ChildItem -Directory | Where-Object { Get-ChildItem $_\*.bak | Where-Object {$_.LastWriteTime -gt $ts} };
$filesToDelete = $dirsToClean | ForEach-Object {Get-ChildItem (Join-Path $_ *) -Include *.bak,*.trn | Where-Object {$_.LastWriteTime -lt $ts}};

变量名应该是不言自明的。

避免这种$ null测试的方法

2 个答案: