Struct contents reference from a non-struct array object.
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I am getting error in following line of code.
if(size(find(cfg.ch1==upper(str(j))),2)==0), valid=0; return; end;
and its called as: if(~checkValidGt(gt)), continue; end
Any solution please??
답변 (1개)
Walter Roberson
2017년 4월 9일
Whatever cfg is, it is not a structure or object
By the way you should consider isempty() instead of what you do with size().
You should consider
if ~any(cfg.ch1 == upper(str(j)))
댓글 수: 2
Asghar Ali
2017년 4월 10일
Walter Roberson
2017년 4월 10일
You cannot reference variables in another script or function.
You can execute a script to have it change the variables in the current workspace, and then you can use that value.
You can can call a function that returns a value and use that value.
You can call a function that returns a structure and assign the structure to a variable and then refer to a field in the variable.
You can define a class that has the field as a property, using the syntax you are already using.
You can define a "package" cfg that has a function ch1 that you could call upon using the syntax you have now.
However there is no way to refer to the inside of a script.
카테고리
도움말 센터 및 File Exchange에서 Variables에 대해 자세히 알아보기
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