我写了一个脚本来将文件复制到“所有用户”桌面或“公共桌面”
但是我们有一个混合的环境。有些人正在使用Windows XP,而其他人正在使用Windows 7。
$SOURCE = "I:\Path\To\Folder\*"
$DESTINATION7 = "c$\Users\Public\Desktop"
$DESTINATIONXP = "c$\Documents and Settings\All Users\Desktop"
$computerlist = Get-Content I:\Path\To\File\computer-list.csv
$results = @()
$filenotthere = @()
$filesremoved = @()
foreach ($computer in $computerlist) {
if((Test-Connection -Cn $computer -BufferSize 16 -Count 1 -ea 0 -quiet))
{
Write-Host "\\$computer\$DESTINATION\"
Copy-Item $SOURCE "\\$computer\$DESTINATION\" -Recurse -force
} else {
$details = @{
Date = get-date
ComputerName = $Computer
Destination = $Destination
}
$results += New-Object PSObject -Property $details
$results | export-csv -Path I:\Path\To\logs\offline.txt -NoTypeInformation -Append
}
}
答案 0 :(得分:4)
DESTINATION是空的。扩展基思的建议:
foreach ($computer in $computerlist) {
if((Test-Connection -Cn $computer -BufferSize 16 -Count 1 -ea 0 -quiet))
{
$OS = Get-WmiObject -Computer $computer -Class Win32_OperatingSystem
if($OS.caption -like '*Windows 7*'){
$DESTINATION = $DESTINATION7
}
if($OS.caption -like '*Windows XP*'){
$DESTINATION = $DESTINATIONXP
}
}
}
这可以避免你得到的错误。 empty $DESTINATION
。
答案 1 :(得分:2)
在您通过$ computerlist的foreach循环中,您可以使用WMI获取每台计算机的操作系统标题:
$OS = Get-WmiObject -Computer $computer -Class Win32_OperatingSystem
Ant然后检查$ OS
if($OS.caption -like '*Windows 7*'){
#Code here for Windows 7
}
#....
答案 2 :(得分:0)
我的目标略有不同......但感谢基础。
del C:\scripts\OS.csv
$computerlist = Get-Content c:\scripts\computerlist.csv
foreach ($computer in $computerlist) {
if((Test-Connection -Cn $computer -BufferSize 16 -Count 1 -ea 0 -quiet))
{
Get-WMIObject Win32_OperatingSystem -ComputerName $computer |
select-object CSName, Caption, CSDVersion, OSType, LastBootUpTime, ProductType| export-csv -Path C:\Scripts\OS.csv -NoTypeInformation -Append
}
}