Fluent validation

Question

Hi Team , .NET core API Fluent Validation Dynamically how set in model class and difference condition is it possible ?any example article or resources please share

Jaimin Shethiya · Answer

Hello Jobin, You mean to say based on the request, automatically identified the fluent validation class and validate that request. This is my understanding correct?

Jaimin Shethiya · Answer

Hello Jobin, Can you more elaborate your requirement so we are understand and give an idea. Thanks

Prasad Raveendran · Answer

Hello Jubin, I would appreciate if you could provide more insights on how you are expecting fluent validation dynamically set on model classes.

Jaimin Shethiya · Answer

Hello Jobin, Hope below code good for you. var employee = new Employee { }; var result = Validate("FluentValidationDemo.Validator.EmployeeValidator", employee); Console.WriteLine($"{nameof(Employee)} validation"); foreach (var item in result.Errors) { Console.WriteLine(item.ErrorMessage); } Console.WriteLine(); Console.WriteLine($"{nameof(Student)} validation"); var student = new Student { }; result = Validate("FluentValidationDemo.Validator.StudentValidator", student); foreach (var item in result.Errors) { Console.WriteLine(item.ErrorMessage); } FluentValidation.Results.ValidationResult Validate(string typeName, object data) { var type = Type.GetType(typeName); object? instance = null; if (type != null) { instance = Activator.CreateInstance(type); } var methods = type.GetMethods(); var method = methods.FirstOrDefault(x => x.IsFinal && x.Name == "Validate"); var response = method.Invoke(instance, [data]); var result = response as FluentValidation.Results.ValidationResult; ArgumentNullException.ThrowIfNull(result); return result; } Output Thanks

Jaimin Shethiya · Answer

Hello Jobin, You need to use a reflection and try to create an object and invoke that method. I used this in my one of the organisation. var type = Type.GetType("Your class name space"); object? instance = null; if (type != null) { instance = Activator.CreateInstance(type, _loggerFactory); } return instance; or instance.MethodName(); Try this way, Hope it will help you. Thanks

Jobin S · Answer

yes jaimin , request validate in dynamically ? any options here fluent validation?

Jobin S · Answer

Dynamically validation set in request model class options here in Fluent validation

Fluent validation

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