I have a survey that has the following requirements

1) Survey is displayed question by question
2) I have a list of 20 brands with an other specify at the end in a multiple choice question
3) I need to ask a respondent about a randomly answered brand for a series of subsequent questions

So for example
B5a Which of the following have you heard of
apples
banana
pears

E1 What do you like about [Random from B5a]

I have looked through the solutions that have been provided in the past but these all include group by group with a hidden text question ... this will not work for me as we have over 100 questions so each would have to be moved into its own group.

Is this even possible?

Thanks in advance

First of all: I assume that the random brand should be chosen from the ones answered, right? So only from the ones they have heard of?

Problem with question by question is where to "hide" the question that creates the randomness.

That's correct Holch ... I have thought the best way to do this would be to create an array in JQuery using the answers given then create a random number from the number of answers given and then use that to select from the array and push that into a single choice question.
Unfortunately my Jquery skills are not that advanced.

Hi,
you can do this with some equations.
I did this some time ago. Maybe not the best, but working.

1. Equation "eqmyString"
I create a string with the "join"-function.
{join(if(Q01_SQ001=="Y","01",""),if(Q01_SQ002=="Y","02",""),if(Q01_SQ003=="Y","03",""),if(Q01_SQ004=="Y","04",""),...)}
You see if the brand was selected the code is added to the string.
You will get something like "0306111417"

2. Equation "eqnBrands"
{count(that.Q01)}
I count the number of entered brands and get "eqnBrands". In this case 5.

3. Equation "eqRandom"
I create a random number (1-numberBrands)
{rand(1,eqnBrands)}
Let's say: 4

4. Equation "eqmyBrand"
I get the substring of my string depending on the random number (eqRandom)
{substr(eqmyString,2*(eqRandom-1),2)}.
In this case substr(eqmyString,6,2) and you get "14".

You have to set (eqRandom-1) because substr starts at position 0.

And in your next question you pipe in the name.?
What do you like about {if(eqmyBrand=="01","Apples",if(eqmyBrand=="02","Bananas",if(eqmyBrand=="03","Pears","")))}?


Best regards
Joffmn

Amazing ... thank you so much ...

Joffm, looks good. But where do you hide those equation type questions in a question by question environment?

Have a look at the sample.

Here it is question by question , and I did it group by group as well.



Best regards
Joffm

Hi there,
I tried to addapt Joffms solution for a question including text as answers
Everything is working for me, exept step one:
"1. Equation "eqmyString"
I create a string with the "join"-function.
{join(if(Q01_SQ001=="Y","01",""),if(Q01_SQ002=="Y","02",""),if(Q01_SQ003=="Y","03",""))}"

My first addaption is looking like this:
{join(if(!is_empty(NW1_SQ001),
if(!is_empty(NW1_SQ002),
if(!is_empty(NW1_SQ003))))}
Here, limesurvey doesn't accept the commas

Afterwards, I tried out an other addaption like this:
{ ... if(!is_empty(NW1_SQ003),3,
if(!is_empty(NW1_SQ002),2,
if(!is_empty(NW1_SQ001),1))))}
This was working yesterday. Today it is only working when I have NW1_SQ018 answered.

Can you help me finding the mistakes?

{join(if(!is_empty(NW1_SQ001),
if(!is_empty(NW1_SQ002),
if(!is_empty(NW1_SQ003))))}
Here, limesurvey doesn't accept the commas

LimeSurvey does accept comma, but your usage of the if-statements seems to be wrong.
We don't see your sample.

Remember:
if(condition,what if true,what if false)

Joffm

Joffm wrote: Remember:
if(condition,what if true,what if false)

Just a correction : OR Since 3.0.2 LimeSurvey version

Thank you very much for your answers.

Remember:
if(condition,what if true,what if false)

that would mean, my example 2 should work properly although it doesn't, right?

We don't see your sample

I also attached my sample for a better comprehensibility of the appearing errors

It would be very nice, if you could find the time for a deeper look inside the code

PS: see question COUNT1