how to construct valid json string from a raw string

Question

Below is MY DTO.

public class CustomerTO
{
  public int CustomerId { get; set;}
  public string CustName { get; set;}
  public string CustJson { get; set;}
}

Below is my LINQ to get Customer records.

var query = (from c in entities.tblCustomers
             select c).toList<CustomerTO>();

This returns the customer collection & in the UI I gets collection as.

Sample:-

    {  
    "CustomerId":113,
    "CustName":"Ram",
    "CustJson":""{\r\n  \"Number\": 143,\r\n  \"IsDeleted\": false,\r\n  \"GapAnalysisChecked\": false,\r\n  \"ShowGraphics\": true,\r\n  \"Impact\": {\r\n    \"Value\": \"DefaultNodeTitle_Impact\",\r\n    \"Details\": null,\r\n    \"DefaultValue\": \"DefaultNodeTitle_Impact\"}
   }

I'm in need to get a valid json string in CustJson varialble.

Please note that in the db, the stored data in CustJSON column is a valid json string.

So I tried this.

foreach(var cust in customers)
{
   if(cust.CustJson != null)
   {
     var parsedJson = JsonConvert.DeserializeObject(cust.CustJson); // this give a valid json
    cust.CustJson  = JsonConvert.SerializeObject(parsedJson);// but this creates a string with \r\n
   }
}

When I try this, in parsed JSON I get the desired JSON. However, when Serialize the parsed JSON again the same string returns.

How do I get the valid json string in Cust.Json??

Is there a better way to get the valid json without the foreach loop?

Answer 1

Uhm, I think I got your problem. You want to include the JSON stored in the CustJson Column into the resulting json, on serializing CustomerTO. Right now this JSON is beeing escaped by the serializer because the serializer doesn't know that this is valid json.

What you are doing now is deserializing the json into a dynmaic object and then serialize it back to a json string into the very same property. Of course this will have no effect on serializing the whole object because the json will simply be escaped again.

You may want to enpower the magic of dynamic a little bit :)

var listWithDeserializedProperty = customers.Select(c => 
{
    dynamic cust = new System.Dynamic.ExpandoObject();
    cust.CustomerId = c.CustomerId;
    cust.CustName = c.CustName;
    if (!string.IsNullOrWitheSpace(c.CustJson))
        cust.CustJson = JsonConvert.DeserializeObject(c.CustJson);
    return cust;
});

return JsonConvert.SerializeObject(listWithDeserializedProperty);

If you now serialize this object, CustJson will be detected as an object and will be included into the resulting json as a whole without beeing escaped again. Of course you could also write a strongly typed view model for this, but for the sake of simplicity...