Zip File Read – pass file name to text field

[Regina]

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

[dmiller]

zip-filename is the value of the File field within the ZipFile Read, so how is that being populated? Is the value specified there or passed in from the previous Snaps?

[Regina]

The value of the File field passed from the previous Snaps. Here is description of each Snap:








Thank you!

[dmiller]

It looks like the value of content-location coming out of the Mapper is the file name minus the path.

If you were to put a Copy after the Mapper, followed by a Mapper to only use content-location (remap to FileName), then join it back at the JSON Splitter.

[Regina]

Thank you, Diane. Unfortunetly, FileName is not coming through after Json Parser and therefore, there is no way to get to it in JSON Splitter.
ZipFile Read produces following output:

JSON Parser follows directly after ZipFile Read and outputs only content of the data file itself:

Therefore, JSON Path does not have FileName:

Any ideas how I can pass FileName from ZipFile Read to JSON Parser and then to JSON Splitter?
Thank you!