JsonParserMultiple 2.1.0

JsonParserMultiple

Convert Json to DataTable without any model or class
Get the result immediately in a DataTable
Create models or classes for convert json in .NET

You can find the sample project in GitHub
This is a simple way to parse your Json result to DataTable.
Instruction of use Json parser multiple :

Methode 1 :

Just follow these simple steps to use JsonParser

• download JsonParserMultiple DLL
• Use below samples code.

Just run this sample code to parse Json to DataTable after download the DLL :

JsonObject

Simple Example :

string JsonStr = "{\"id\":\"1\",\"name\":\"Jack\",\"age\":\"36\"}";
JsonParser.JParser JP = new JsonParser.JParser();
DataTable dt = JP.JsonConvert(JsonStr);

JsonArray

Example 1 :

string JsonStr = "[{\"id\":\"1\",\"name\":\"Jack\",\"age\":\"36\"}
                  ,{\"id\":\"2\",\"name\":\"Janatan\",\"age\":\"30\"}
                  ,{\"id\":\"3\",\"name\":\"Natasha\",\"age\":\"32\"}]";
JsonParser.JParser JP = new JsonParser.JParser();
DataTable dt = JP.JsonConvert(JsonStr);

Example 2 :

It may that our JsonArray have different values in their jsonobjects.
For example :

string JsonStr = "[{\"id\":\"1\",\"name\":\"Jack\",\"age\":\"36\"}
                  ,{\"id\":\"2\",\"name\":\"Janatan\",\"age\":\"30\"}
                  ,{\"id\":\"3\",\"name\":\"Natasha\",\"zipcode\":\"5896472358\"}]";
JsonParser.JParser JP = new JsonParser.JParser();
DataTable dt = JP.JsonConvert(JsonStr);

We have age in first and second jsonobjects, but in third one we have zipcode instead of age.
Consequently in two first row we have zipcode with empty value and in third row we have empty value for age column and this value '5896472358' for zipcode column.
Like below table :

Object In Object

Sometimes one of our value consists a JsonObject even a JsonArray. we can use JsonParser for value of that column specifically

Example 1 :

string JsonStr = "{\"id\":\"1\",\"product\":\"TV\",\"number\":\"1\"
                  ,\"extrainfo\":{\"product_id\":\"157895004\",\"product_serial\":\"8874500\",\"product_model\":\"M23TU\"}}";

JsonParser.JParser JP = new JsonParser.JParser();
DataTable dt = JP.JsonConvert(JsonStr);

In this JsonObject that concern to a sale order, extrainfo include of another JsonObject.
If you need extrainfo values like ' product_id , product_serial , ... ' you must pass value of extrainfo column to JsonConvert.
The result is like this below table in first step of convert.

In the next step I pass value of extrainfo column to JsonConvert and get values.
Pass value like below code

DataTable dt_extrainfo = JP.JsonConvert(dt.Rows[0]["extrainfo"]);

Note: You can use loop (for , while...) for convert another rows simply

DataTable dt_extrainfo;
for (int i = 0; i < dt.Rows.Count; i++)
{
    dt_extrainfo = new DataTable();
    dt_extrainfo = JP.JsonConvert(dt.Rows[i]["extrainfo"].ToString());
}

Example 2 :

Note: Another solution for this case is JsonConvertToDictionary method.

In previous example we had extrainfo column that must convert, we used again JsonConvert and get values in another DataTable
now we can receive our information in a Dictionary. It is the DataTable that extrainfo column has a JasonObject
and we want to get valus such as product_serial, pay attention to below code.

Dictionary<string, string> DicExtrainfo = JP.JsonConvertToDictionary(dt.Rows[i]["extrainfo"].ToString());

Convert to Dictionary

You can get the result in a Dictionary instead of DataTable

Note: This method can uses just for one row not more
☑ {"product_id":"157895004","product_serial":"8874500","product_model":"M23TU"} (True)
☒ [{"id":"1","product":"refrigerator"},{"id":"2","product":"carpet"}] (False) we have two rows in this case

Example 1 :

string JsonStr = "{\"id\":\"1\",\"name\":\"Natasha\",\"zipcode\":\"5896472358\"}";
JsonParser.JParser JP = new JsonParser.JParser();
Dictionary<string, string> Dic = JP.JsonConvertToDictionary(JsonStr);

JsonConvertToDictionary could use like JsonConvert.

Object/Array In Object

Example 1 :

In this JsonObject that concern to a sale order, extrainfo include of another JsonObject.
If you need extrainfo values like ' product_id , product_serial , ... ' you must pass value of extrainfo key to JsonConvertToDictionary.

string JsonStr = "{\"id\":\"1\",\"product\":\"TV\",\"number\":\"1\"
                  ,\"extrainfo\":{\"product_id\":\"157895004\",\"product_serial\":\"8874500\"
                  ,\"product_model\":\"M23TU\"}}";

JsonParser.JParser JP = new JsonParser.JParser();
Dictionary<string, string> Dic = JP.JsonConvertToDictionary(JsonStr);
Dictionary<string, string> DicExtrainfo = JP.JsonConvertToDictionary(Dic["extrainfo"]);

Methode 2 :

Parse Json individually step by step by JsonObject and JsonArray classes.

Example 1 :

string str = "{\"Table1\":[{\"id\":\"1\",\"name\":\"Koorosh\",\"age\":\"36\"},{\"id\":\"2\",\"name\":\"Ahoora\",\"age\":\"30\"}]
 ,\"Table2\":[{\"id\":\"1\",\"product\":\"Speacker\",\"price\":\"60\"},{\"id\":\"2\",\"product\":\"Monitor\",\"price\":\"250\"}]}";
JsonParser.JsonObject jsonobject = new JsonParser.JsonObject(str, "Table2");
string array_str = jsonobject.GetJsonArray();
JsonParser.JsonArray jarray = new JsonParser.JsonArray(array_str);
List<string> results = jarray.ConvertJsonArray();
for (int i = 0; i < results.Count; i++)
{
      Dictionary<string, string> jobj = jsonobject.GetJsonObject(results[i]);
      string id = jobj["id"];
      string product = jobj["product"];
      string price= jobj["price"];
}

There is a JsonObject that includes two JsonArrays that challed 'Table1' and 'Table2'
We want to parse 'Table2', so use above code
1 ) First and second line : get JsonArray that called 'Table2'
2 ) Third and fourth line : get JsonObjects in JsonArray
3 ) Then by use a simple loop we can access to our values

Note: Don't forget each of the values can includes another JasonObject or a JsonArray.