PowerShell本地磁盘属性没有输出

时间:2013-12-12 11:39:07

标签: properties powershell-v3.0 psobject

我正在为磁盘构建报告,但需要帮助。这是代码:

$outputs = @()
$disks = get-wmiobject -class "Win32_LogicalDisk" -Filter 'driveType=3' -namespace "root\CIMV2"
foreach ( $disk in $disks ) {
    $output = New-Object PSObject -Property @{
        deviceID     = $disk.caption
        FileSystem   = $disk.fileSystem
        FreeSpace    = $disk.freeSpace/1GB
        Size         = $disk.size/1GB
        VolumeName   = $disk.volumeName

        used = "{0:N1}" -f ((("{0:N1}" -f ($disk.size/1gb)).ToString().replace(",",".") - ("{0:N1}" -f ($disk.freespace/1gb)).ToString().replace(",","."))   /   ("{0:N1}" -f ($disk.size/1gb)).ToString().replace(",",".") * 100)

        used2 = $used
    }
    $outputs += $output
}
$outputs | select deviceID, VolumeName, size, used, used2  | Format-Table

一切都很好,但变量used2

deviceID VolumeName             Size used used2
-------- ----------             ---- ---- -----
C:       win        49,1992149353027 87,4
D:       data       170,701168060303 70,7

1 个答案:

答案 0 :(得分:0)

$used在该范围内不存在为变量,因此used2设置为$null

您需要在调用$used之前创建变量New-Object 。此外,在计算使用的百分比时,你正在做太多格式化 - 将格式保存到最后。下面,我做了两个更改

$used = "{0:P1}" -f (($disk.size - $disk.FreeSpace) / $disk.size);
$output = New-Object PSObject -Property @{
        deviceID     = $disk.caption
        FileSystem   = $disk.fileSystem
        FreeSpace    = $disk.freeSpace/1GB
        Size         = $disk.size/1GB
        VolumeName   = $disk.volumeName
        used = $used
        used2 = $used
        }