更新时间:2021-11-30 19:18:07
我已经进行了测试,以成功获取文件名.
I've made a test to get the FileNames successfully.
我创建了与您相同的文件结构.
I created the same file structure as yours.
在GetMetaData1活动中,让我们定义根文件夹 2020Folder
的数据集.然后,我们使用 Child Items
来获取所有子文件夹.
At the GetMetaData1 activity, let's define a DataSet of the root folder 2020Folder
. Then we use Child Items
to get all the subfolders.
@activity('Get Metadata1').output.childItems
列出文件夹列表.@activity('Get Metadata1').output.childItems
to foreach the Folder list.
@item().name
,以将子文件夹名称传递给GetMetadata2活动.然后,我们可以使用GetMetadata2活动从子文件夹中获取 Child Items
.@item().name
to pass the subfolder name to the GetMetadata2 activity. Then we can use the GetMetadata2 activity to get the Child Items
from the subfolder.
@activity('Get Metadata2').output.childItems [0]
来获取文件名.@activity('Get Metadata2').output.childItems[0]
to get the filename.
我们可以看到数组的输出.
The output we can see the array.