在powershell $ x.FullName中没有返回完整路径

Question

我在下面有一个powershell脚本,它接受一个配置文件并删除与正则表达式匹配的x天以前的文件.

配置文件:

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

不过这是输出:

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

我期望输出包含完整的文件路径,因为我正在使用.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为什么我没有获得路径的全名(C:\ logs)？

谢谢Brad

$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.您没有在导致"将被删除"的条件下使用FullName.更改:

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

至:

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

要回答您的注释,如果要在字符串中展开的变量上调用属性,则必须将其包围$(),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