Powershell：基于动态菜单的程序来复制文件

Question

这样做的目的是将一个（或多个）文件从一个用户选择的动态位置复制到另一个用户选择的动态位置。

Write-Host "Choose where to copy FROM"

MainFolder = (Get-ChildItem "C:\Users$env:USERNAME\FROM" -Directory | Sort-Object)
$menu = @{}
for ($i=1;$i -le MainFolder.count; $i++) {
    Write-Host "$i. $(MainFolder[$i-1].name)"
    $menu.Add($i,(MainFolder[$i-1].name))
    }

[int]$ans = Read-Host 'Enter selection'
$selection = $menu.Item($ans)
# The directory after the selection is only one, however it's different for every user; this remains the same for "COPY TO"
$First = (Get-ChildItem "C:\Users$env:USERNAME\FROM$selection" -Directory)

# This directory is the same for every user; this remains the same for "COPY TO"
$DataFROM = (Set-Location $First\USERFILE) 

Clear-Host


Write-Host "Choose where to copy TO"

MainFolder = (Get-ChildItem "C:\Users$env:USERNAME\TO" -Directory | Sort-Object)
$menu2 = @{}
for ($i=1;$i -le MainFolder.count; $i++) {
    Write-Host "$i. $(MainFolder[$i-1].name)"
    $menu2.Add($i,(MainFolder[$i-1].name))
    }

[int]$ans2 = Read-Host 'Enter selection'
$selection2 = $menu2.Item($ans2)
$Second = (Get-ChildItem "C:\Users$env:USERNAME\TO$selection2" -Directory)

$DataTO = (Set-Location $Second\USERFILE)

Clear-Host

第一节和第二节到此结束。第三部分允许用户选择他们想要复制的文件——这是一个单独的功能，效果很好，但与此相关时会出错。我已经简化了函数以获得相同的错误：

# The Select-Object section being the simplified part...
Get-ChildItem $DataFROM | Select-Object "DATA.log*"  | Copy-Item -Destination $DataTO

我希望发生的是将数据从初始动态位置复制到最终动态目标。

我收到一条错误消息，指出“无法将参数绑定到参数 'Path'，因为它是一个空字符串。”它指向 $DataTO 变量。

这导致我忽略第三部分并测试我发现的变量，如果我在前两部分之后使用以下内容：


Get-ChildItem $DataFROM

# or

Get-ChildItem $DataTO

它 returns 从上次选择的菜单返回信息，完全忽略变量。那么动态菜单是不是不能这样用，还是我用错了菜单？

有人能告诉我哪里误入歧途了吗？

Answer 1

更新：自从 re-worked 以来，我已经将其更新为满足我要求的需求的以下内容：（希望这将帮助其他需要类似内容的人）


[array]$sMainFolder = Get-ChildItem "C:\Users$env:USERNAME\DATA-TO_FROM" -Directory | ForEach-Object {
          Get-ChildItem $_.FullName | ForEach-Object {
              $found = Get-ChildItem $_.FullName | Where-Object { $_.Name -eq 'USERFILE' } | Select-Object -First 1
              if ($found) {
                  [PSCustomObject]@{
                      RootFolder = [string]$_.FullName
                      Path = $found.FullName
                  }
              }
          }
      }

      [array]$dMainFolder = Get-ChildItem "C:\Users$env:USERNAME\DATA-TO_FROM" -Directory | ForEach-Object {
        $subs = Get-ChildItem $_.FullName
        $found = $subs | ForEach-Object { Get-ChildItem $_.FullName } | Where-Object { $_.Name -eq 'USERFILE' } | Select-Object -First 1
            if ($found) {
                [PSCustomObject]@{
                    RootFolder = [string]$_.FullName
                    Path = $found.FullName
                }
            }
        }

function ShowMenu ($Folders, $Message) {
    Write-Host $Message
    for ($i = 1; $i -le $Folders.count; $i++) {
        $name = $Folders[$i - 1].RootFolder
        Write-Host "$i. $name"
    }
    [int]$ans = Read-Host 'Enter Selection'
    Return $Folders[$ans -1].Path
}

$DataFROM = ShowMenu $sMainFolder "Choose where to copy FROM"
$DataTO = ShowMenu $dMainFolder "Choose where to copy TO"

Clear-Host

Copy-Item "$DataFROM\*" $DataTO -Recurse -Force

Powershell：基于动态菜单的程序来复制文件

Powershell: Dynamic Menu-based program to copy file

powershell

menu

copy-item