Use expression as a variable parameter

[ijulias]

Hi everybody,

I've defined the variable "vArrodoniment" like this:

if($1>20,num($1,'##.##0'),if($1>1,num($1,'##.##0,0'),num($1,'##.##0,00')))

So, it has different correct results:

=$(vArrodoniment(0.05)) ---> 0,05
=$(vArrodoniment(10)) ---> 10,0
=$(vArrodoniment(22)) ---> 22

Now, I would like to pass a parameter as a result of a function instead of a hardcoded number.

I've the expression:

=$(=Only({1<Codi_Indicador={'83'}>} formula_municipi_any))

It's ok and it results 71,12.

So I would like to use this expression as the parameter of the variable vArrodoniment.

I use it inside like this.

=$(vArrodoniment($(=Only({1<Codi_Indicador={'83'}>} formula_municipi_any))))

And there is the next error "Error: Error in expression: '}' expected ".

Anybody knows if it's possible to use the result of an expression as a parameter of a variable?

Thanks a lot!

[MarkWillems · Accepted solution]

I got this to work by putting the Only() function in it's own hidden expression within a straight table and then referencing it within the variable function call, which sits within it's own expression:

=$(vArrodoniment("Hidden Expression name"))

Putting the expression directly within the call to the variable would not give me an error, but it would also sum all the figures in a table for each row in the table and ignore the dimensions entirely.

[ijulias]

Thanks MarkWillems,

I've several cases I wanted to use the same solutions; some of them I've got the same result as you, and another are so difficult because every execution of the variable Qv try to execute it in a context that sometimes is not the desired.

At the end, I've used the pass variable parameter in one case.