XSD - define type with parameter

StackOverflow https://stackoverflow.com/questions/22090814

  •  18-10-2022
  •  | 
  •  

Вопрос

Is it possible to define parameter for a type ? I want to have type that represents fixed length string but I dont want do define multiple such types for different lengths, that is TFixedString(N) so I could use TFixedString(2) and TFixedString(16).

Это было полезно?

Решение

No you can not do this. It is not supported in XSD.

Лицензировано под: CC-BY-SA с атрибуция
Не связан с StackOverflow
scroll top