How to Access list of JSON objects

Question

I am trying to get the "name" and "value" attribute from the JSON file which i am loading externally.

I tried accessing the list Valuestore and I am returned with an error.

"""Cannot deserialize the current JSON object (e.g. {"name":"value"}) into type 'System.Collections.Generic.List`1[ExternalJSON.Form1+Valuestore]' because the type requires a JSON array """"

How can i access the list value store and print the name and value in the listview.

My Code

    public class Value
    {
        public string value { get; set; }
    }

    public class Valuestore
    {
        public string name { get; set; }
        public string type { get; set; }
        public Value value { get; set; }
    }

    public class RootObject
    {
        public string version { get; set; }
        public List<Valuestore> Valuestore { get; set; }
    }

    private void button1_Click(object sender, EventArgs e)
    {

        var json = File.ReadAllText(".....\\static_settings");

        var settingsNamelst1 = Newtonsoft.Json.JsonConvert.DeserializeObject<RootObject>(json);
        var settingsNamelst2 = Newtonsoft.Json.JsonConvert.DeserializeObject<List<Valuestore>>(json);

        foreach (var lst in settingsNamelst2)
        {
            listView1.Items.Add(lst.name);
        }

    }

And my JSON file is

{
    "version": "2",
    "valuestore": [
        {
            "name": "abcd1",
            "type": "string",
            "value": {
                "value": "0002"
            }
        },
        {
            "name": "abcd2",
            "type": "string",
            "value": {
                "value": "001"
            }
        }
    ]
}

and I am returned with an error. what error? Include it in your question. — Matt Burland
– Matt Burland, Commented Feb 18, 2015 at 15:01
The error is caused by Newtonsoft.Json.JsonConvert.DeserializeObject<List<Valuestore>>(json);. Remove that line and use what @EZI suggested instead — Andrew Whitaker
– Andrew Whitaker, Commented Feb 18, 2015 at 15:07

Nikola.Lukovic · Accepted Answer · 2015-02-18 15:17:34Z

2

The easiest thing you can do is to deserialize it dynamically, and acces the objects over dynamic properties.

var json = @"{
                        ""version"": ""2"",
                        ""valuestore"": [
                            {
                                ""name"": ""abcd1"",
                                ""type"": ""string"",
                                ""value"": {
                                    ""value"": ""0002""
                                }
                            },
                            {
                                ""name"": ""abcd2"",
                                ""type"": ""string"",
                                ""value"": {
                                    ""value"": ""001""
                                }
                            }
                        ]
                    }";

        dynamic jsonTemp = JsonConvert.DeserializeObject(json);
        foreach (var i in jsonTemp.valuestore)
        {
            Console.WriteLine("name: {0}, value: {1} \n", i.name, i.value.value);
        }

answered Feb 18, 2015 at 15:17

Nikola.Lukovic

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1 Comment

raj3209812 Over a year ago

Selecting this answer. Crisp and clear solution I was looking for.

Yuval Itzchakov · Accepted Answer · 2015-02-18 15:07:39Z

0

Your settingsNamelst1 property is already deserialized and contains the Valuestore, no need to de-serialize it again:

var settingsNamelst1 = JsonConvert.DeserializeObject<RootObject>(json);
foreach (var item in settingsNamelst1.Valuestore)
{
    listView1.Items.Add(item.name);
}

The reason you're seeing the exception is because your JSON schema doesn't match List<Valuestore>, it matches your RootObject.

answered Feb 18, 2015 at 15:07

Yuval Itzchakov

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4 Comments

raj3209812 Over a year ago

Thank you very much for your explanation. How can i access value attribute. For example var settingsNamelst1 = JsonConvert.DeserializeObject<list<value>.(json); Because it give the same error for value

EZI Over a year ago

@raj3209812 Come on, don't be so lazy. VS will help you

raj3209812 Over a year ago

Ok. Got the hint from Nikola.Lukovic answer.

raj3209812 Over a year ago

value.value made the difference. Thanks every one

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