Regina
7 years agoNew Contributor II
Zip File Read – pass file name to text field
Hello, we have a data coming in a zip file. One of the requirements is to read file name and send it to a text field so it can be loaded in to one of the columns in Oracle table.
For example:
File name is SV_4YEVwN5KjncCtVj.zip
Output result (Oracle table):
Below is a pipeline I have:
ZipFileRead produces following result.
How can I extract “SV_4YEVwN5KjncCtVj” from the "zip-filename": "file:///C:/Users/SVC_SN~1/AppData/Local/Temp/sl_pipe_bdca1c58-dd15-403e-9b86-5b6c38102d1b/SV_4YEVwN5KjncCtVj.zip"
and pass it to a JSON Splitter?
Any suggestions would be greatly appreciated.
Thank you!
Regina