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Asked: 2016-07-01 18:23:25 +0800 CST

如果匹配powershell，比较两个项目

  • 772

我在文本文件中导出驱动器名称电影。

Get-WmiObject win32_logicaldisk| foreach-object {$volumename -Match "movies" } | select movies > c:\nn.txt

后来比较它但不成功总是得到错误的结果有人可以帮忙吗？

$test=Get-WmiObject win32_logicaldisk| foreach-object {$volumename -Match "movies" } | select movies > c:\nn.txt
    $array =get-content c:\nn.txt
    $real= Get-WmiObject win32_logicaldisk| foreach-object {$volumename -Match "movies"} | select movies
    If ($array -eq $real)

    {write-host "it was success" -foregroundcolor blue}
    Else {write-host "it was fail" -foregroundcolor green}
powershell

1 个回答

  1. Best Answer
    get-wmiobject win32_logicaldisk | ? Volumename -eq "movies" | select -ExpandProperty Volumename | out-file "c:\nn.txt"
$array =get-content  "c:\nn.txt"
$real =  get-wmiobject win32_logicaldisk | ? Volumename -eq "movies" | select -ExpandProperty Volumename
If ($array -eq $real)

{write-host "it was success" -foregroundcolor blue}
Else {write-host "it was fail" -foregroundcolor green}
    • 0

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