Question

我正在尝试将一批文件（文件名以6位开头的文件）从temp文件夹复制到永久位置，但不包括新位置中已经存在的文件。

复制完成后，我想将文件名和复制文件的新路径导出到CSV中。

获取旧文件位置并导出为CSV很容易，我不确定如何获取新文件位置。

我的脚本如下：

# Prompt for file origin
$file_location = Read-Host -Prompt "Where do you want your files to come from?"
# Prompt for file destination 
$file_destination = Read-Host -Prompt "Where do you want your files to go? `n(They won't be copied if they're already there)"

# Save contents of file destination - used to check for duplicates
$dest_contents = Get-ChildItem $file_destination

<# For all of the files of interest (those whose names begin with 6 digits) in $file_location, 
     determine whether that filename already exists in the target directory ($file_destination)
     If it doesn't, copy the file to the destination
     Then save the filename and the new** filepath to a CSV #>

Get-ChildItem -Path $file_location |      
    Where-Object { $_.Name -match '^\d{6}' -And !($dest_contents -Match $_.Name ) } |
    Copy-Item -Destination $file_destination -PassThru | 
    Select-Object -Property Name, FullName |             # **Note: This saves the old filepath, not the new one
    Export-CSV -Path "$file_location\files_copied.csv" -NoClobber

Answer 1

您可以对代码进行一些更改

# Prompt for file origin
$file_location = Read-Host -Prompt "Where do you want your files to come from?"
# Prompt for file destination 
$file_destination = Read-Host -Prompt "Where do you want your files to go? `n(They won't be copied if they're already there)"

# Save contents of file destination - used to check for duplicates
$dest_contents = Get-ChildItem $file_destination | Select-Object -ExpandProperty Name

<# For all of the files of interest (those whose names begin with 6 digits) in $file_location, 
     determine whether that filename already exists in the target directory ($file_destination)
     If it doesn't, copy the file to the destination
     Then save the filename and the new** filepath to a CSV #>

Get-ChildItem -Path $file_location |      
    Where-Object { $_.Name -match '^\d{6}' -and ($dest_contents -notcontains $_.Name ) } |
    ForEach-Object { 
        Copy-Item -Path $_.FullName -Destination $file_destination 
        # emit a PSObject storing the Name and (destination) FullName of the file that has been copied
        # This will be used to generate the output in the 'files_copied.csv'
        New-Object -TypeName PSObject -Property ([ordered]@{ Name = $_.Name; FullName = (Join-Path $file_destination $_.Name)})
    } | 
    Export-CSV -Path "$file_location\files_copied.csv" -NoTypeInformation -Force

请注意，我只收集目标路径中已经存在的文件的名称，而不是fileInfo对象。这使它变得“精简”得多，因为收集的唯一原因是要收集要比较的文件名。

现在，您已经为“ files_copied.csv”使用了固定的名称，我个人认为，通过在其上添加当前日期（例如）来使其更通用，是一个好主意

Export-CSV -Path ("{0}\files_copied_{1}.csv" -f $file_location, (Get-Date).ToString("yyyy-MM-dd")) -NoTypeInformation -Force

P.s。我在这里使用[ordered]，因此输出将始终具有相同顺序的属性。但是，这需要PowerShell v3或更高版本。另外，如果您需要确保代码仅复制文件而不是目录，则建议查看-File命令上的-Attributes或Get-ChildItem开关。如果您的PowerShell版本是2.0，则可以使用Where-Object { !$_.PSIsContainer }构造仅过滤掉文件。

Answer 2

我将使用另一种方法测试迭代源文件时是否存在目标。
要附加到当前的csv（日志）文件中，您需要删除新文件的标题。
由于Get-ChildItem允许范围[0-9]，因此您可以直接选择前导数字

## Q:\Test\2018\08\08\SO_51738853.ps1
# Prompt for file origin
$SrcDir = Read-Host -Prompt "Where do you want your files to come from?"

# Prompt for file destination 
$DstDir = Read-Host -Prompt "Where do you want your files to go? `n(They won't be copied if they're already there)"

$SrcFiles = Join-Path (Get-Item $SrcDir).Fullname [0-9][0-9][0-9][0-9][0-9][0-9]*

$CopiedFiles = ForEach ($SrcFile in (Get-ChildItem -Path $SrcFiles)){
    $DstFile = Join-Path $DstDir $SrcFile.Name
    If (!(Test-Path $DstFile)){
        $SrcFile | Copy-Item  -Destination $DstFile -PassThru | 
            Select-Object -Property Name, FullName
    }
}    

# Check if log file exists, if yes append, otherwise export.
$LogFile = Join-Path $SrcDir "files_copied.csv"
If ($CopiedFiles){
    If (!(Test-Path $LogFile)){
        Export-CSV -Path $LogFile -InputObject $CopiedFiles -NoTypeInformation
    } else {
        # We need to remove the Header appending to the present csv
        $CopiedFiles | ConvertTo-Csv -NoType | Select-Object -Skip 1 | Add-Content -Path $LogFile
    }
}

如何复制文件，然后将目标路径保存到CSV

2 个答案: