Receiver side file name same as Sender side file name

Question

Hi,
  We want to pick a file from one location and put into another location using XI <u><b>with the same name in which the file was lying there in the source directory/folder</b></u>. This we need to do it w/o opening the file and hence we cannot use the variable substitution also.
Any clue????
Thanks in advance
Jayakrishnan

Answer 1

Hi,
There was a similar discussion which went on this topic.
Retain file name
In my opinion, we can easily achieve this using modules. The steps are very simple.
(any file) ->[module #1]-> XML -> Sender CC
The XML produced will have 2 tags say,
<filename></filename><data_base64></data_base64>
where fileame is the name of the file and data is the base64 converted data. Similarly
Receiver CC -> XML -> [module #2] ->  (any file)
During the second step there will be a default step module  called localejbs/CallSapAdapter. Just remove this entry and have only your module entry provided the module #2 must implement all FileOutputStream functions to write the file from the stream, extracting the filename from the xml node. Since you are removing the CallSapAdapter module, the default file which you give in the Receiver CC is not created, instead the filename which comes alone the tag is created. I hope this information helps you.
I am planning to put a blog on this
Best regards,
Felix