Hi team
My problem statement is to take input numbers and seperate by comma post 16 digit however I am using this expression but not able to excute
Can some help me in resolving the expression Plzz
String.Join(“,”,Enumerable.Range(0,CInt(InputText.Length/16)+1).Select(Function (x) String.Join(“”,InputText.Skip(x*16).Take(16).toarray)).toArray)
When I excute I am getting error as end of expression expected.
Request you to please help me resolving the same
ppr
(Peter Preuss)
August 27, 2020, 3:25pm
2
I tested the statement and did not get a validation error. can you share screenshot or XAML with us. Thanks
@ppr
Kindly find the screenshot
ppr
(Peter Preuss)
August 27, 2020, 3:28pm
4
plase share in a form that we can see the entire statement
@ppr
Error ERROR Validation Error Compiler error(s) encountered processing expression ““output”,String.Join(”,“,Enumerable.Range(0,CInt(InputText.Length/16)+1)(Select(Function (x) String.Join(”“,InputText.Skip(x*16).Take(16).toarray)).toArray)”.
End of expression expected.
ppr
(Peter Preuss)
August 27, 2020, 3:48pm
7
give a try:
remove the term output or do a correct concatenation of “output” + Str…
@ppr I am getting error object reference not set to instance of an object as
ppr
(Peter Preuss)
August 27, 2020, 4:13pm
9
I guess you are trying something like this the input box:
SliceStringBySegmentSize_V2.xaml (5.5 KB)
For reference not set issues go for debuging and inspect the values of variables. Also check the definition scopes
@ppr
I am getting Something like this which is shown in the screenshot
However the scope of definition is crt it is defined to entire sequence.
ppr
(Peter Preuss)
August 27, 2020, 5:17pm
11
i assume that in the variable panel Input… is the Variable InputText that you are using for the statement. Here you can directly see thatg it is null and cause the issue. A valid value is to assign to the ariable before using it.
Yeah it is taking as null
Soo how do I assign variable
ppr
(Peter Preuss)
August 27, 2020, 5:22pm
13
this ping pong is not effectively. Can you share your XAML?
How to share call??could u help me if I copy it is just copying error with expression
@ppr
Kndly find the xaml file attachedSequence1.xaml (4.6 KB)
ppr
(Peter Preuss)
August 28, 2020, 2:46pm
17
find reviewed XAML:
Sequence1_V2.xaml (5.7 KB)
the wrong variable was used and so the entered data from first input dialog was ignored / not used.
Also have a look on the approach with options as dropdown