Dynamic File name generation

[Anonymous]

Can anyone suggest an approach to build a job which will create a file with with the name same as the source file. As I am new to this tool I am not sure from where shall I start. 

any help is highly appreciated.

[manodwhb]

@saurabh06101987,can you chekc below link.

https://www.youtube.com/watch?v=JFzk5s55048

[Anonymous]

Hi Manohar, 

i appreciate your effort but i guess that one is for SSIS,solution in  talend is what i am looking for . 

[manodwhb]

@saurabh06101987,did you checked below link?

https://community.talend.com/t5/Archive/Generate-dynamic-files/td-p/181921

[Anonymous · Accepted solution]

Just use a tfileList (the directory where your file is present) and in the fliemask you can give depending on your file name. Connect iterate link from tfileList to tfileInputDelimited (in directory, give variable ((String)globalMap.get("tFileList_1_CURRENT_FILEPATH"))

In your tfileoutputDelimited component you can give value for directory as your output directory + ((String)globalMap.get("tFileList_1_CURRENT_FILE")) -----this variable stores the current filename from tFileList component.

I hope this helps.

[Anonymous]

Hey Ravinderk,

This worked, thanks a lot for your time.