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abhi90
Creator II
Creator II

Create Separate Output File based on Input File Name Identifier

Hi All,

I am trying to implement Encryption Decryption of Context File. I am able to do that. Now challenge is I have 11 DataSources. So 11 Context File is there. I am trying to create a Single Job which will run at first and will encrypt all the Context Files. Requirement is if any Context File is having Name as File_A.txt then encrypted Context File will generate with File_A.txt. Can it be achiveble? @TRF@rhall@vboppudi@manodwhb

 

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5 Replies
abhi90
Creator II
Creator II
Author

Hi,

 

Can Anyone please help @nthampi

TRF
Champion II
Champion II

Yes it can!
What's the question?
Anonymous
Not applicable

Hi,

 

    There is a custom component in Talend exchange called tFileEncryptPGP. Could you please try it?

 

https://exchange.talend.com/#marketplaceproductoverview:marketplace=marketplace%252F1&p=marketplace%...

 

Warm Regards,

 

Nikhil Thampi

     

 

abhi90
Creator II
Creator II
Author

Hi @TRF,

My Question is How to generate Output File based on my Input FileName Pattern. For E.G if I have 11 Files in a Location and iterating it in a FileList I want to generate a Output .txt File for each Pattern Matching Up with Input FileName. So if My Source FileNames are Abhijit_File1_DS1.txt,Abhijit_File1_DS2.txt then Output File will be generating based on the _DS value present. So If DS1 is xxx and DS2 will be yyy File1 will be having Name as ----_xxx.txt.

TRF
Champion II
Champion II

Connect a tJava after tFileList and use the String.split() method to get the parts of the tFileList_1_CURRENT_FILE and generate the desired filename for each output file. The result should be stored into a global or context variable.