Best way to get the information of a json file and use it

Question

I'm building an app in c# and i need to parse / create / edit / delete inforamtions (etc...) in a .json file. Underneeth is how the file .json will look like.

I was thinking of doing dynamic arrow of class (don't know if it is the way we say it in english). Is there a faster and simpler way to do this?

Thank you very much!

{
  "name": [
   {
     "var1": string or int,
     "var2": string or int,
     etc..
     .
     .
     .
   },
   {
     "var1": string or int,
     "var2": string or int,
     etc..
     .
     .
     .
   }
 ]
}

Newtonsoft JSON should help you here. https://www.newtonsoft.com/json — Matt Oestreich, Jan 10 '19 at 14:31
If the data that you get is known you just need to use the [Serialize and Deserialize](https://www.newtonsoft.com/json/help/html/SerializingJSON.htm) strategy. — andre, Jan 10 '19 at 14:37

TanvirArjel · Answer 1 · 2019-01-12T18:47:50.607

2

First make two classes as follows:

public class Name
{
   public List<SampleObject> SampleObjects {get; set;}
}

public class SampleObject
{
    public int var1;
    public int var2;
   ....
}

Then try as follows:

public void JsonFileReader()
{
    using (StreamReader r = new StreamReader("yourJsonFile.json"))
    {
        string json = r.ReadToEnd();
        List<Name> items = JsonConvert.DeserializeObject<Name>(json);
    }
}

Note: Don't forget to adding using Newtonsoft.Json

edited Jan 12 '19 at 18:47

answered Jan 10 '19 at 14:34

TanvirArjel

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Note that this makes use of the [Json.NET](https://www.newtonsoft.com/json) library. – user247702 Jan 10 '19 at 14:36
Yes! Thanks for mentioning this! – TanvirArjel Jan 10 '19 at 14:36
You are missing a class: the original JSON is a single object containing a property `name` which contains an array of `Name` objects. If you try to deserialize the JSON to a `List` it will fail. – Brian Rogers Jan 12 '19 at 18:33
@BrianRogers Please check! I have updated my answer! – TanvirArjel Jan 12 '19 at 18:49
Construct classes and deserialize [like this](https://dotnetfiddle.net/IbqfYj) – Brian Rogers Jan 12 '19 at 18:55

Best way to get the information of a json file and use it

1 Answers1