@DavidBrabant是正确的，将通配符放在路径中是正常的，但是我认为你可以在这种情况下逃脱它，因为你通过foreach声明来管理结果。

@DavidBrabant is right that it is not normal to put the wildcard in the path, but I think you can get away with it in this instance as you are piping the results through a foreach statement.

我认为问题是你的第一个参数 CreateFromDirectory 应该是一个目录名，但你已经传递了一个文件名。使用变量名称（$ Source和$ source）确实没有帮助。

I believe the problem is that your first parameter of CreateFromDirectory should be a directory name, but you have passed it a filename. This isn't really helped by the use of variable names ($Source and $source).

当你打电话给 CreateFromDirectory 它将包含第一个zip文件的全名，因为以下行：

When you call CreateFromDirectory it will contain the full name to the first zip file because of the following line:

$Source = $_.fullName

我猜你有一个过滤器'* .txt'你要添加单个文件而不是整个文件夹到一个zip文件然后它更多涉及。请参阅此处以获取示例： http：// msdn。 microsoft.com/en-us/library/hh485720(v=vs.110).aspx

I'm guessing that as you have a filter '*.txt' you want to add individual files rather than the whole folder to a zip file then it is a little more involved. See here for an example: http://msdn.microsoft.com/en-us/library/hh485720(v=vs.110).aspx

但是如果你只是想压缩文件夹然后使用：

But if you simply want to zip the folder then use:

$sourceFolder = "C:\folder1"
$destinationZip = "c:\zipped.zip" 
[Reflection.Assembly]::LoadWithPartialName( "System.IO.Compression.FileSystem" )
[System.IO.Compression.ZipFile]::CreateFromDirectory($sourceFolder, $destinationZip)