我正在使用脚本根据EXIF数据重命名文件。
param([string]$path)
# http://blogs.technet.com/b/jamesone/archive/2007/07/13/exploring-photographic-exif-data-using-powershell-of-course.aspx
[reflection.assembly]::loadfile( "C:\Windows\Microsoft.NET\Framework\v2.0.50727\System.Drawing.dll") | out-null
function MakeString {
$s=""
for ($i=0 ; $i -le $args[0].value.length; $i ++) {
$s = $s + [char]$args[0].value[$i]
}
return $s
}
$files = Get-ChildItem -Path $path
foreach ($file in $files) {
if ($file.Extension -ne ".jpg") { continue }
if ($file.Name -match "^(\d+)-(\d+)-(\d+)") { continue }
$exif = New-Object -TypeName system.drawing.bitmap -ArgumentList $file.FullName
$captureDate = MakeString $exif.GetPropertyItem(36867)
$captureDate = ($captureDate -replace ":", '-').Substring(0,19)
$newFilename = $captureDate + " " + $file.Name.Trim()
$file.Name + " -> " + $newFilename
$file |Rename-Item -NewName $newFilename
}
从EXIF读取日期没问题,但是当我尝试重命名文件时,我收到以下错误消息:
Rename-Item : The process cannot access the file because it is being used by another process.
At D:\Norwegen\RenamePhotos.ps1:25 char:12
+ $file |Rename-Item -NewName $newFilename
+ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
+ CategoryInfo : WriteError: (D:\Norwegen\P1270465 (1920x1433).jpg:String) [Rename-Item], IOException
+ FullyQualifiedErrorId : RenameItemIOError,Microsoft.PowerShell.Commands.RenameItemCommand
当我监控ProcMon中的目录时,我可以看到文件关闭的时间很晚:
(查看突出显示的行,它来自当前正在处理的文件条目中间的早期文件)
那么,如何关闭文件(可能仍然从EXIF读取打开)所以我可以重命名它?
我已经尝试关闭打开的文件:
Remove-Variable exif
$file.Close()
答案 0 :(得分:3)
由于您只是从文件中读取,因此调用Dispose()
应该可以解决问题。实施例
foreach ($file in $files) {
if ($file.Extension -ne ".jpg") { continue }
if ($file.Name -match "^(\d+)-(\d+)-(\d+)") { continue }
$exif = New-Object -TypeName system.drawing.bitmap -ArgumentList $file.FullName
$captureDate = MakeString $exif.GetPropertyItem(36867)
#Disposing object as soon as possible
$exif.Dispose()
$captureDate = ($captureDate -replace ":", '-').Substring(0,19)
$newFilename = $captureDate + " " + $file.Name.Trim()
$file.Name + " -> " + $newFilename
$file |Rename-Item -NewName $newFilename
}