2 ACCEPTED SOLUTIONS
This finds the location of "BEG", slices off the rest of the string, splits the remainder into an array, and returns the 4th index.
Because "BEG" might appear elsewhere in the string, this one may be more reliable if the start of the "BEG" segment is always formatted as "~BEG*"
Here's the formula as text so you can copy/paste. Just replace both "x" variables with the data pill.
x.slice(x.index("~BEG*")..-1).split('*')[3]
I don't think all of this could be done in a single line (at least not by me!), so I wrote a quick code snippet you can use in a Ruby action. It will output the following:
- begIDs: an array of BEG ID values
- begSegmentsFinal: an array of updated BEG segments after replacing *RL* with *RE* -- this is mostly for reference, but I included it anyway
- updatedInput: this is the entire string input updated after replacing *RL* with *RE*
Here's the actual output:
Set up a Ruby action with one string input called "string", and pass in your string.
Set up the output like this:
Use this code:
string = input[:string]
begSegments = string.scan(/(~BEG\*\d{2}\*[A-Z]{2}\*\d+\*\*)/).flatten
begIds = []
begSegmentsFinal = []
updatedInput = ''
begSegments.each { | s |
begIds.push(s.split('*')[3])
begSegmentUpdated = s.include?('*RL*') ? s.gsub('*RL*','*RE*') : s
begSegmentsFinal.push(begSegmentUpdated)
updatedInput = string.gsub(s, begSegmentUpdated)
}
{
begIds: begIds,
# if you want a comma-delimited list, just use begIds.join(',') but you'll also have to update the output schema so begIds is a string and not an array
begSegmentsFinal: begSegmentsFinal,
updatedInput: updatedInput
}
It first uses a regex to find the BEG segments up to the ID value. Then it loops through the array of matches and uses gsub to replace RL/RE. In my opinion, this is a bit more selective/careful because it's only doing it on a portion of the BEG segment and not the entire input string where other instances of *RL* might exist, but I have no idea.
There are probably better or more efficient ways of doing this, but this should work until it doesn't!
I don't think all of this could be done in a single line (at least not by me!), so I wrote a quick code snippet you can use in a Ruby action. It will output the following:
- begIDs: an array of BEG ID values
- begSegmentsFinal: an array of updated BEG segments after replacing *RL* with *RE* -- this is mostly for reference, but I included it anyway
- updatedInput: this is the entire string input updated after replacing *RL* with *RE*
Here's the actual output:
Set up a Ruby action with one string input called "string", and pass in your string.
Set up the output like this:
Use this code:
string = input[:string]
begSegments = string.scan(/(~BEG\*\d{2}\*[A-Z]{2}\*\d+\*\*)/).flatten
begIds = []
begSegmentsFinal = []
updatedInput = ''
begSegments.each { | s |
begIds.push(s.split('*')[3])
begSegmentUpdated = s.include?('*RL*') ? s.gsub('*RL*','*RE*') : s
begSegmentsFinal.push(begSegmentUpdated)
updatedInput = string.gsub(s, begSegmentUpdated)
}
{
begIds: begIds,
# if you want a comma-delimited list, just use begIds.join(',') but you'll also have to update the output schema so begIds is a string and not an array
begSegmentsFinal: begSegmentsFinal,
updatedInput: updatedInput
}
It first uses a regex to find the BEG segments up to the ID value. Then it loops through the array of matches and uses gsub to replace RL/RE. In my opinion, this is a bit more selective/careful because it's only doing it on a portion of the BEG segment and not the entire input string where other instances of *RL* might exist, but I have no idea.
There are probably better or more efficient ways of doing this, but this should work until it doesn't!
