如何配置“复制文件”和“移动文件”文件系统任务中的操作？
查看MSDN 文件系统任务编辑器（常规页面）< /a> 文章，我不清楚应该为源值和目标值提供什么。
对于移动文件，它表示“移动文件时，请勿在您提供的目标目录路径中包含文件名。”，但对于则不然复制文件。
我本以为我可以提供
SourceVariable = User::MyFileName
DestinationVariable = User::MyDestinationFolder
我对移动或复制文件没有任何乐趣。 MyFileName var 应在 ForEach 容器中使用 *.ext 设置，但从监视窗口中查看，它似乎被设置为 filename.ext< /code> 而不是 c:\source\filename.ext。
在包的前面部分，我使用相同的逻辑来打开和读取文件，并且工作正常。这里可能有什么问题？
Looking at the MSDN File System Task Editor (General Page) article, it is not clear to me what should be supplied for the source and target values.
It says "When moving a file, do not include a file name in the directory path that you provide as the destination." for move file, but not for copy file.
I would've thought that I could supply
SourceVariable = User::MyFileName
DestinationVariable = User::MyDestinationFolder
I'm not having any joy with either move or copy file. The MyFileName var should be set during a ForEach container using *.ext, but from looking in the watch window, it appears to be set to filename.ext and not c:\source\filename.ext.
In an earlier part of the package, I'm using the same logic to open and read the files and it's working fine. What could be the issue here?
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我没有注意到 foreach 容器上的单选按钮，特别是我没有设置“完全合格”。
我仍然不明白为什么我的包的“文件打开”部分工作得很好，因为它似乎没有使用该路径。
I had failed to notice the radio buttons on the foreach container, specifically I did not have 'fully qualified' set.
I still don't understand why the 'file opening' part of my package is working perfectly, given that it does not appear to use the path.