powershell - 在 powershell $x.FullName 不返回完整路径

Question

我在下面有一个 powershell 脚本，它需要一个配置文件并删除与正则表达式匹配的 x 天之前的文件。

配置文件:

path,pattern,days,testrun
C:\logs\,^data_access_listener.log,7,false

但是，这是输出:

Would have deleted 000a19f6-a982-4f77-88be-ca9cc51a2bcbuu_data_access_listener.log
Would have deleted 00189746-2d46-4cdd-a5bb-6fed4bee25a7uu_data_access_listener.log

我希望输出包含完整的文件路径，因为我使用的是 .FullName 属性，所以我希望输出如下:

Would have deleted C:\logs\000a19f6-a982-4f77-88be-ca9cc51a2bcbuu_data_access_listener.log
Would have deleted C:\logs\00189746-2d46-4cdd-a5bb-6fed4bee25a7uu_data_access_listener.log

如果我使用 $x.FullName 为什么我没有得到路径的全名 (C:\logs)？

谢谢
布拉德

$LogFile = "C:\deletefiles.log"
$Config = import-csv -path C:\config.txt

function DeleteFiles ([string]$path, [string]$pattern, [int]$days, [string]$testrun){
    $a =  Get-ChildItem $path -recurse | where-object {$_.Name -notmatch $pattern}    
    foreach($x in $a) {
    $y = ((Get-Date) - $x.LastWriteTime).Days
        if ($y -gt $days -and $x.PsISContainer -ne $True) {

            if ($testrun -eq "false") {
                write-output “Deleted" $x.FullName >>$LogFile 
            } else {
                write-output “Would have deleted $x” >>$LogFile 
            }


        }
    }
}

foreach ($line in $Config) {
    $path = $line.path
    $pattern = $line.pattern
    $days = $line.days
    $testrun = $line.testrun

    DeleteFiles $path $pattern $days
}

Answer 1

看起来像一个错字。在导致“Would have Deleted”的情况下，您没有使用 FullName。改变:

write-output “Would have deleted $x” >>$LogFile

到:

write-output “Would have deleted " $x.FullName >>$LogFile

要回答您的评论，如果您想调用在字符串中展开的变量的属性，则必须将其括在 $() 中。 ，以便解析器知道调用整个表达式。像这样:

write-output “Would have deleted $($x.FullName)" >>$LogFile