Question

如果文件重复，则该脚本可在复制文件时重命名文件。我需要先重命名当前目标文件，然后按原样复制源文件。有什么想法吗？

function Copy-FilesWithVersioning{
    Param(
        [string]$source,
        [string]$destination
    )
    Get-ChildItem -Path $source -File
        ForEach-Object {
            $destinationFile = Join-Path $destination $file.Name
            if ($f = Get-Item $destinationFile -EA 0) {
                # loop for number goes here
                $i = 1
                $newname = $f.Name -replace $f.BaseName, "$($f.BaseName)_$I")
                Rename-Item $destinationFile $newName
            }
            Copy-Item $_ $destination
        }
}

Copy-FilesWithVersioning c:\scripts\Source c:\scripts\DestinationA

错误：

At line:10 char:53
+             if($f = Get-Item $destinationFile -EA 0){
+                                                     ~
Missing closing '}' in statement block or type definition.
At line:8 char:23
+         ForEach-Object{
+                       ~
Missing closing '}' in statement block or type definition.
At line:2 char:34
+ function Copy-FilesWithVersioning{
+                                  ~
Missing closing '}' in statement block or type definition.
At line:13 char:77
+ ...         $newname = $f.Name -replace $f.BaseName, "$($f.BaseName)_$I")
+                                                                         ~
Unexpected token ')' in expression or statement.
At line:15 char:13
+             }
+             ~
Unexpected token '}' in expression or statement.
At line:17 char:9
+         }
+         ~
Unexpected token '}' in expression or statement.
At line:18 char:1
+ }
+ ~
Unexpected token '}' in expression or statement.
    + CategoryInfo          : ParserError: (:) [], ParentContainsErrorRecordException
    + FullyQualifiedErrorId : MissingEndCurlyBrace

Answer 1

您看到的错误是由以下行中的假括弧引起的：

$newname = $f.Name -replace $f.BaseName, "$($f.BaseName)_$I")

从行尾删除括号，这些错误将消失。

尽管您的代码中还有其他几个错误，所以即使修复了该错误，代码仍然无法正常工作。

您缺少Get-ChildItem和ForEach-Object之间的管道。将一个cmdlet的输出传递到另一个cmdlet时需要使用此
。
```
Get-ChildItem -Path $source -File |
    ForEach-Object {
        ...
    }
```
变量$file未定义。在PowerShell管道中，您要使用“当前对象”变量（$_）。更改此行
```
$destinationFile = Join-Path $destination $file.Name
```
进入
```
$destinationFile = Join-Path $destination $_.Name
```
$_
```
Copy-Item $_ $destination
```
被扩展为仅文件名，而不是完整路径。将其更改为
```
Copy-Item $_.FullName $destination
```
更好的是，将Copy-Item语句移到{{1}之后之后，这样就不必首先明确指定源（cmdlet从管道）：
```
ForEach-Object
```
请注意，您必须将当前对象输出回管道，并指定目的地作为命名参数（Get-ChildItem ... | ForEach-Object { ... $_ # need this line to pass the current object back into the pipeline } | Copy-Item -Destination $destination），以便后者工作。
您检查目标文件夹中是否存在文件有点尴尬。请改用-Destination $destination。您可以从当前对象构造新文件名。
```
Test-Path
```

Answer 2

尝试通过以下链接提供代码：https://www.pdq.com/blog/copy-individual-files-and-rename-duplicates/：

type RequireAllPropertiesContractResolver() =
    inherit CamelCasePropertyNamesContractResolver()

    override __.CreateProperty(memb, serialization) =
        let prop = base.CreateProperty(memb, serialization)
        let isRequired = not (prop.PropertyType.IsGenericType && prop.PropertyType.GetGenericTypeDefinition() = typedefof<option<_>>)
        if isRequired then prop.Required <- Required.Always
        prop

重命名文件（如果已存在）

2 个答案: