Reputation: 11
I am new to C#
Following is the JSON string I am getting from the web API.
I am trying to store the JSON string into a class and then store the JSON string into an SQL table.
But the C# code is failing to deserialize JSON into class. And the message box returns the null exception error.
JSON
{
"Count":3,
"data":[
{
"Cost1":{
"amount":111,
"currencyCode":"ABC"
},
"Cost2":{
"amount":22.2,
"currencyCode":"XYZ"
},
"Id":"007"
},
{
"Cost1":{
"amount":555,
"currencyCode":"ABC"
},
"Cost2":{
"amount":444,
"currencyCode":"XYZ"
},
"Id":"008"
},
{
"Cost1":{
"amount":666,
"currencyCode":"ABC"
},
"Cost2":{
"amount":8882,
"currencyCode":"XYZ"
},
"Id":"009"
}
],
"pending":[
],
"@up":"Test Data"
}
C# Code
public class ParceJSN {
public int Count {
get;
set;
}
public string data {
get;
set;
}
public string pending {
get;
set;
}
public string up {
get;
set;
}
}
public void Main() {
Task < string > task = MakeRequest(db_token); //Returns the JSON string
var fr = task.Result;
ParceJSN rst = JsonConvert.DeserializeObject < ParceJSN > (fr.ToString());
MessageBox.Show(rst.TotalCount.ToString());
Dts.TaskResult = (int) ScriptResults.Success;
}
Upvotes: 1
Views: 1930
Reputation: 388
Your C# model is incorrect. Here's a really handy tool that I use when I need to generate C# classes based on some JSON - json2csharp.com
The correct class is:
public class Cost1 {
public int amount { get; set; }
public string currencyCode { get; set; }
}
public class Cost2 {
public double amount { get; set; }
public string currencyCode { get; set; }
}
public class Datum {
public Cost1 Cost1 { get; set; }
public Cost2 Cost2 { get; set; }
public string Id { get; set; }
}
public class Root {
public int Count { get; set; }
public List<Datum> data { get; set; }
public List<object> pending { get; set; }
[JsonProperty("@up")]
public string Up { get; set; }
}
Now that you have the correct model, you can now do the following to deserialize your JSON string into a Root object which is defined above:
Root myDeserializedClass = JsonConvert.DeserializeObject<Root>(YOUR_JSON_STRING);
Upvotes: 1