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0

I want my model to accept all 3 different types of value in id(basically Guid and int only)

This is my model-

public class Base
 
 public dynamic Id get; set; 
 public string Type get; set; 
 

This is my json-

 "id": "31367556-cda7-4fce-9d8a-2cd6f46544f9",
 "type": "form"
,


 "id": "123",
 "type": "form"
,


 "id": 456,
 "type": "form"

c# .net json serialization json.net

share|improve this question

asked Mar 22 at 5:41

tushar parashartushar parashar

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Probably a duplicate of this stackoverflow.com/a/42334643/224370 with the added Guid option.
  
  – Ian Mercer
  Mar 22 at 6:10
  

  
  
  
  

add a comment | 

0

I want my model to accept all 3 different types of value in id(basically Guid and int only)

This is my model-

public class Base
 
 public dynamic Id get; set; 
 public string Type get; set; 
 

This is my json-

 "id": "31367556-cda7-4fce-9d8a-2cd6f46544f9",
 "type": "form"
,


 "id": "123",
 "type": "form"
,


 "id": 456,
 "type": "form"

c# .net json serialization json.net

share|improve this question

asked Mar 22 at 5:41

tushar parashartushar parashar

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Probably a duplicate of this stackoverflow.com/a/42334643/224370 with the added Guid option.
  
  – Ian Mercer
  Mar 22 at 6:10
  

  
  
  
  

add a comment | 

I want my model to accept all 3 different types of value in id(basically Guid and int only)

This is my model-

public class Base
 
 public dynamic Id get; set; 
 public string Type get; set; 
 

This is my json-

 "id": "31367556-cda7-4fce-9d8a-2cd6f46544f9",
 "type": "form"
,


 "id": "123",
 "type": "form"
,


 "id": 456,
 "type": "form"

c# .net json serialization json.net

share|improve this question

asked Mar 22 at 5:41

tushar parashartushar parashar

1

public class Base
 
 public dynamic Id get; set; 
 public string Type get; set; 
 

 "id": "31367556-cda7-4fce-9d8a-2cd6f46544f9",
 "type": "form"
,


 "id": "123",
 "type": "form"
,


 "id": 456,
 "type": "form"

share|improve this question

asked Mar 22 at 5:41

tushar parashartushar parashar

1

share|improve this question

asked Mar 22 at 5:41

tushar parashartushar parashar

1

asked Mar 22 at 5:41

tushar parashartushar parashar

1

asked Mar 22 at 5:41

tushar parashartushar parashar

1

1 Answer
1

active

oldest

votes

0

First of all you need to answer on the following questions: why you really need dynamic type? How are you going to use it?. After that you can choose approach to implement. Maybe you need just string.

Another approach, is to use to 2 properties and for example enum which indicate for you which id should be selected.

Make your Data model which accept string

public class Base
 
 public string Id get; set; 
 public string Type get; set; 
 

And after you received and parse it you need to map this Data Transfer Model to Model and use it in your Application only Model types.

Mapping can be

//your ApplicationModel
public class BaseModel

 public Guid GuidIdget;set;
 public int Idget;set;
 public IdTypeget;set;

 public enum IdType
 
 Guid,
 Int
 


public BaseModel Map(Base dto)

 var model=new BaseModelType= dto.Type;
 var guidResult = Guid.TryParse(dto.Id,out model.GuidId);
 if(!guidResult) int.TryParse(dto.Id, out model.Id);
 model.IdType = guidResult?IdType.Guid:IdType.Int;
 return model

using dynamic type for me is Bad practice. I know only few cases where it acceptable.
More info you can find here:Is the use of dynamic considered a bad practice?

share|improve this answer

edited Mar 22 at 7:01

answered Mar 22 at 6:53

Yauhen SampirYauhen Sampir

22329

add a comment | 

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0

First of all you need to answer on the following questions: why you really need dynamic type? How are you going to use it?. After that you can choose approach to implement. Maybe you need just string.

Another approach, is to use to 2 properties and for example enum which indicate for you which id should be selected.

Make your Data model which accept string

public class Base
 
 public string Id get; set; 
 public string Type get; set; 
 

And after you received and parse it you need to map this Data Transfer Model to Model and use it in your Application only Model types.

Mapping can be

//your ApplicationModel
public class BaseModel

 public Guid GuidIdget;set;
 public int Idget;set;
 public IdTypeget;set;

 public enum IdType
 
 Guid,
 Int
 


public BaseModel Map(Base dto)

 var model=new BaseModelType= dto.Type;
 var guidResult = Guid.TryParse(dto.Id,out model.GuidId);
 if(!guidResult) int.TryParse(dto.Id, out model.Id);
 model.IdType = guidResult?IdType.Guid:IdType.Int;
 return model

using dynamic type for me is Bad practice. I know only few cases where it acceptable.
More info you can find here:Is the use of dynamic considered a bad practice?

share|improve this answer

edited Mar 22 at 7:01

answered Mar 22 at 6:53

Yauhen SampirYauhen Sampir

22329

add a comment | 

0

First of all you need to answer on the following questions: why you really need dynamic type? How are you going to use it?. After that you can choose approach to implement. Maybe you need just string.

Another approach, is to use to 2 properties and for example enum which indicate for you which id should be selected.

Make your Data model which accept string

public class Base
 
 public string Id get; set; 
 public string Type get; set; 
 

And after you received and parse it you need to map this Data Transfer Model to Model and use it in your Application only Model types.

Mapping can be

//your ApplicationModel
public class BaseModel

 public Guid GuidIdget;set;
 public int Idget;set;
 public IdTypeget;set;

 public enum IdType
 
 Guid,
 Int
 


public BaseModel Map(Base dto)

 var model=new BaseModelType= dto.Type;
 var guidResult = Guid.TryParse(dto.Id,out model.GuidId);
 if(!guidResult) int.TryParse(dto.Id, out model.Id);
 model.IdType = guidResult?IdType.Guid:IdType.Int;
 return model

using dynamic type for me is Bad practice. I know only few cases where it acceptable.
More info you can find here:Is the use of dynamic considered a bad practice?

share|improve this answer

edited Mar 22 at 7:01

answered Mar 22 at 6:53

Yauhen SampirYauhen Sampir

22329

add a comment | 

First of all you need to answer on the following questions: why you really need dynamic type? How are you going to use it?. After that you can choose approach to implement. Maybe you need just string.

Another approach, is to use to 2 properties and for example enum which indicate for you which id should be selected.

Make your Data model which accept string

public class Base
 
 public string Id get; set; 
 public string Type get; set; 
 

And after you received and parse it you need to map this Data Transfer Model to Model and use it in your Application only Model types.

Mapping can be

//your ApplicationModel
public class BaseModel

 public Guid GuidIdget;set;
 public int Idget;set;
 public IdTypeget;set;

 public enum IdType
 
 Guid,
 Int
 


public BaseModel Map(Base dto)

 var model=new BaseModelType= dto.Type;
 var guidResult = Guid.TryParse(dto.Id,out model.GuidId);
 if(!guidResult) int.TryParse(dto.Id, out model.Id);
 model.IdType = guidResult?IdType.Guid:IdType.Int;
 return model

using dynamic type for me is Bad practice. I know only few cases where it acceptable.
More info you can find here:Is the use of dynamic considered a bad practice?

share|improve this answer

edited Mar 22 at 7:01

answered Mar 22 at 6:53

Yauhen SampirYauhen Sampir

22329

public class Base
 
 public string Id get; set; 
 public string Type get; set; 
 

//your ApplicationModel
public class BaseModel

 public Guid GuidIdget;set;
 public int Idget;set;
 public IdTypeget;set;

 public enum IdType
 
 Guid,
 Int
 


public BaseModel Map(Base dto)

 var model=new BaseModelType= dto.Type;
 var guidResult = Guid.TryParse(dto.Id,out model.GuidId);
 if(!guidResult) int.TryParse(dto.Id, out model.Id);
 model.IdType = guidResult?IdType.Guid:IdType.Int;
 return model

share|improve this answer

edited Mar 22 at 7:01

answered Mar 22 at 6:53

Yauhen SampirYauhen Sampir

22329

answered Mar 22 at 6:53

Yauhen SampirYauhen Sampir

22329

answered Mar 22 at 6:53

Yauhen SampirYauhen Sampir

22329