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:
Fil...
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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!
Correct, it would not come through this way. My apologies for not including an image early. My intent was to collect the information earlier in the pipeline, Copy it out to a second branch, then join it back in with a Join or Union, whichever is appropriate for the data.
As @tstack said, there are several ways to do the same thing.
Thank you, Diane and Tim! All information is very helpful
