Using tips gleaned from this, this, and this, I've finally been able to get a series of file backup scripts going. However, there's one little thing that I've been unable to solve. No runtime errors, but when I run this script,
$originalPath = "\\Server\Path\_testData\"
$backupPath = "\\Server\Path\_backup\"
#
function supportBackup
{
"$($originalPath) copying DOC XLS PPT JPG GIF PDF WAV AVI to $($backupPath)"
Get-ChildItem $originalPath\* -Include *.doc*, *.xls*, *.ppt*, *.jpg, *.gif, *.pdf, *.wav, *.avi | `
foreach {
$targetFile = $backupPath + $_.FullName.SubString($originalPath.Length);
New-Item -ItemType File -Path $targetFile -Force;
Copy-Item $_.FullName -destination $targetFile
}
"Support File Backup Completed"
}
supportBackup
The original file path gets dumped into the destination directory instead of just the files.
What I want:
\\Server\Path\_backup\files-from-testData-directory
What I get:
\\Server\Path\_backup\_testData\files-from-testData-directory
I know the problem is closely related (if not identical) to this question, but after studying it and trying to apply some of the wisdom from there, using various iterations of the $_.Name
variables, I realize I don't have as good an understanding as I thought I did. I need someone to explain to me HOW the destination path and filename are being constructed with the given variables, and what alternate variables (or code) I need to use to achieve my desired results. There's something that's not clicking for me and I need help understanding it.