Setting a flexible Model in Web API for Get controller

Question

I am developing a simple Web API with only GET controller. Currently,I have the following model, which is the return type of my GET controller:

namespace _1WebApplication.Models
{
    [DataContract]
    public class SystemStatus
    {        
        public string SystemName { get; set; }       
        public string Good { get; set; }      
        public string Bad { get; set; }        
        public string MachineStatus { get; set; }
    }
}

and my GET api looks like this:

// GET api/values
public SystemStatus Get()
{
    //Read the data from XML file here
    string systemName = "REW_1";
    string good = "60";
    string bad = "10";
    string machineStatus = "Running";

    var SystemStatus = new SystemStatus
    {
        SystemName = systemName,
        Good = goodCount,
        Bad = badCount,
        MachineStatus = machineStatus
    };

    return SmartscanStatus;    
}

QUESTION: In the above case, I can return the values of only 4 variables (i.e. SystemName , Good , Bad and MachineStatus ). But actually, I want to read the parameters and values from an XML file and then return those values.

For example, I have this XML file

<SystemName>mysystem</SystemName>
<Good>60</Good>
<Bad>10</Bad>
<MachineStatus>stop</MachineStatus>
<IpAddress>127.0.0.1</IpAddress>        
<Username>username</Username>       
<Password>password</Password>

In this case, the XML file has some more variable but since my model is fixed, I am not able to return them. So, how to make my model flexible/dynamic so that I can decide the variable names and values from an XML file.

score -1 · Answer 1 · answered Oct 05 '18 at 15:22

-1

Read the XML with XDocument and convert it into dynamic object

ExpandoObject provides a way to construct dynamic properties.

public SystemStatus Get()
{
    XDocument xmlDoc = XDocument.Parse(xmlData); 
    string jsonStr = JsonConvert.SerializeXNode(xmlDoc);
    dynamic dynamicObject = JsonConvert.DeserializeObject<ExpandoObject>(jsonStr);

    return dynamicObject;
}

answered Oct 05 '18 at 15:22

Hary

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But where should I define the return type i.e. `SystemStatus`? in Models? and how? – skm Oct 08 '18 at 07:33
ExpandoObject will be creating that for you – Hary Oct 08 '18 at 08:03
so, you mean that I don't have to make any changes in the Models? – skm Oct 08 '18 at 08:05
It is not going to use `SystemStatus` model. Did you check `ExpandoObject`? – Hary Oct 08 '18 at 08:06
It throws a runtime error at the `return` statement. – skm Oct 08 '18 at 08:50

Setting a flexible Model in Web API for Get controller

1 Answers1