Is join(separator, question) supported yet?

[DougBrowning]

I found the join(separator, question) function listed on this help page

https://doc.arcgis.com/en/survey123/desktop/create-surveys/xlsformformulas.htm

But when I use it I only get the value in the last repeat.  No comma or other repeat data at all.  It changes to the last value in the repeat whenever a repeat is added.

I saw some posts of it not being ready yet but that was back from Nov 2017.

Thanks

[JamesTedrick]

Hi Doug,

The function join() is working properly for me in this sample- can you check to see what the difference might be?

[DougBrowning · Accepted solution]

My code is the same as yours.  But your sheet is doing the same thing for me.  occ_names will just display the value of the repeat I am on.  So if on repeat 1 it just lists repeat 1 value.  If I move to repeat 2 I get just repeat value 2.  It changes as I cycle back through the repeats also.

I have Connect 3.0.128.   ...And so I went and got 3.2.196 and now it works!

Thanks

[TKSHEP]

Hello,

Is there by chance a way to accomplish both of these command in a repeat?  

For example I have a repeat to select_one that are names and I am trying to get the name to show separately.  I am trying to get the related table to show separately in Dashboard.  I hope this makes sense. 

oice-name(choice, 'question')	Used for select_one and select_multiple questions. Returns the label associated with the name of the choice in the given question. Be aware the question must be defined inside quotes.	jr:choice-name(maybe, '${question_one}')
join(separator, question)	Concatenates all answers to a given question in a repeat, separated by the given separator.	join(',', ${question_in_repeat})

Thank you!

[DougBrowning]

Not sure combining like this would work but prob not

join(',',jr:choice-name(maybe, '${question_one}'))

So just add a second hidden field in the repeat that gets the label

question2 calculate of jr:choice-name(maybe, '${question_one}')

Then do your join on that hidden field.

join(',',${question2})

hope that helps.