How to add a new property to an existing json without defining model class in C#?

Question

{
    "version": "15.1",
    "count": 8,
    "data": [
        {
            "furit_items": [
                {
                    "price": 12.6,
                    "base_price": 6.59,
                    "text": "banana",
                    "product_id": "1234",
                    "product_name": "banana",
                    "quantity": 1
                }
            ],
            "product_sub_total": 12.6,
            "product_total": 12.6,
            "shipments": []
        }
    ]
}

This is my json data getting through a service.

Is this possible to add a new property {"quality":"good"} under "furit_items" object?

What is class of this object?

Atilla Ozgur
– Atilla Ozgur

2015-06-25 06:30:01 +00:00
Commented Jun 25, 2015 at 6:30 — Atilla Ozgur
– Atilla Ozgur, Commented Jun 25, 2015 at 6:30
Don't have class . I am getting this json through service.

Jero
– Jero

2015-06-25 06:34:35 +00:00
Commented Jun 25, 2015 at 6:34 — Jero
– Jero, Commented Jun 25, 2015 at 6:34

Sajal · Accepted Answer · 2015-06-25 06:44:51Z

3

Try this:

dynamic original = JsonConvert.DeserializeObject(json, typeof(object));
original.data[0].furit_items[0].quality = good;
var modifiedJson = JsonConvert.SerializeObject(original , Formatting.Indented);

answered Jun 25, 2015 at 6:44

Sajal

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7 Comments

Revils Over a year ago

I have tried this, and it works for me. (Put quotes on the good, "good")

Jero Over a year ago

Error : Cannot perform runtime binding on a null reference

Jero Over a year ago

original.data[0].furit_items[0].quality = "good"; - this what I have checked

Sajal Over a year ago

The name of the your JSON has to be replaced with json in the first line of the answer. Did you do that?

Jero Over a year ago

dynamic original = JsonConvert.DeserializeObject(result, typeof(object)); original.data[0].furit_items[0].quality = "good"; var modifiedJson = JsonConvert.SerializeObject(original, Formatting.Indented);

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Behzad · Accepted Answer · 2015-06-25 06:53:21Z

1

why you do not want to define the model class ? i totally recommend it, since whatever you do become more safe and easy. http://json2csharp.com/ simply generate the model for you.

if you do not want to use the model at least use json.net - jobject.

answered Jun 25, 2015 at 6:53

Behzad

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2 Comments

Jero Over a year ago

couldn't define modal class because it is a dynamic json data

Behzad Over a year ago

so how can you find the path of adding a new property ?

Akshita · Accepted Answer · 2015-06-25 09:12:44Z

1

you can use Linq to Json of Json.Net as

JObject obj = JObject.Parse(jsontext);
obj["new_prop"] = "value";//new property as per hirarchy ,same for replacing values
string newjson=obj.ToString();

answered Jun 25, 2015 at 9:12

Akshita

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Comments

Youku KMSFan · Accepted Answer · 2015-06-25 07:13:16Z

0

You can Try like this,it is easy for using javascript code.

data.furit_items.field={"quality":"good"}

answered Jun 25, 2015 at 7:13

Youku KMSFan

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