如何正确获取变量名并使用powershell将其导入到任务计划中？

Question

I have some problem with this line of code. Currently I'm able to loop all the xml files name inside a folder. But the how to use that variable and put after C:\\Windows\\System32\\Tasks\\Job\\? Currently the powershell always detect it as the text.

$fileDirectory = "C:\Windows\System32\Tasks\Job\*.xml";
foreach($file in Get-ChildItem $fileDirectory)
{
    $file.name
Register-ScheduledTask -xml (Get-Content "C:\Windows\System32\Tasks\Job\$file.name" | Out-String) -TaskName $file.name -TaskPath "\Job" -User "MyAccount" –Force
}

Answer 1

You need to surround your $file.name in "C:\\Windows\\System32\\Tasks\\Job\\$file.name" with the Subexpression Operator $() , otherwise it's just trying to substitute the $file part.

Compare this:

PS> $file = get-item "c:\temp\temp.txt"
PS> "C:\Windows\System32\Tasks\Job\$file.name"
C:\Windows\System32\Tasks\Job\C:\temp\temp.txt.name

with this:

PS> $file = get-item "c:\temp\temp.txt"
PS> "C:\Windows\System32\Tasks\Job\$($file.name)"
C:\Windows\System32\Tasks\Job\temp.txt

The first example evaluates $file (which is a System.IO.FileInfo object) to C:\\temp\\temp.txt and just replaces $file in the string, leaving a trailing .name as literal text.

The second example evaluates $($file.name) to temp.txt (which is a string) instead and replaces the entire $($file.name) subexpression.

But simpler still, in your case you could just use $file.FullName which gives the full path:

PS> $file = get-item "c:\temp\temp.txt"
PS> $file.FullName
c:\temp\temp.txt

Answer 2

Another option is to use the f-format operator.

$fileDirectory = "C:\Windows\System32\Tasks\Job\*.xml";
foreach($file in Get-ChildItem $fileDirectory)
{
    $file.name
Register-ScheduledTask -xml (Get-Content "C:\Windows\System32\Tasks\Job\{0}" -f $file.name | Out-String) -TaskName $file.name -TaskPath "\Job" -User "MyAccount" –Force
}