Parse JSON String into List

Question

It is needed to parse JSONString into List. (List of instances) I'm trying to use JSON.NET by Newtonsoft.

I have classes:

    public class Item
    {
        public int ID { get; set; }
        public string Name { get; set; }
        public string Url { get; set; }
        public string Manufactorers { get; set; }

    }

The JSON string looks something like this:

[
{     
 "Column0":23.0,
 "Column1":"Евроен",
 "Column2":"https://www.123.com",
 "Column3":"Фак"
},
{
 "Column0":24.0,
 "Column1":"Еил",
 "Column2":"https://www.123.com",
 "Column3":"Старт"
}
]

I've been trying to do something like this:

string JSONString = string.Empty;
JSONString = JsonConvert.SerializeObject(result);
List<Item> items = JsonConvert.DeserializeObject<List<Item>>(JSONString);

But it returns 0 and null. I have no idea, how to fix it.

Also here I truy to parse Excel file. This code works, but after deserialization, I have just 0 and null.

var filePath = @"..\..\..\..\doc.xlsx";

using (var steam = File.Open(filePath, FileMode.Open, FileAccess.Read))
{
   using (var reader = ExcelReaderFactory.CreateReader(steam))
   {
       var result = reader.AsDataSet().Tables["Лист1"];

       string JSONString = string.Empty;
       JSONString = JsonConvert.SerializeObject(result);

       List<Item> items = JsonConvert.DeserializeObject<List<Item>>(JSONString);

    }
}

This is not C, is it perhaps C#? Please pay attention when selecting tags. — Some programmer dude
– Some programmer dude, Commented Feb 24, 2020 at 9:30
There are no properties named Column0, Column1... in your class. That's why the Deserialization fails — Cid
– Cid, Commented Feb 24, 2020 at 9:34
If I add properties named Column0, Column1, I can see this error Newtonsoft.Json.JsonReaderException: 'Could not convert string to integer: ID. Path '[0].Column0', line 1, position 16.' — Yulia Koval
– Yulia Koval, Commented Feb 24, 2020 at 9:36
result is var result = reader.AsDataSet().Tables["Лист1"];.This is part of the parsing Excel file. — Yulia Koval
– Yulia Koval, Commented Feb 24, 2020 at 9:38

Tomas Chabada · Accepted Answer · 2020-02-24 10:06:47Z

1

The naming of JSON and your class does not match. This can be fixed using JsonProperty attributes:

[JsonProperty("Column0")]
public decimal ID { get; set; }

Second, JSON deserizlizer can not deserialize string "23.0" to int when there is decimal point. You can retype ID to decimal or double to make it work.

Little test here:

public class TestClass
{
    [JsonProperty("Column0")]
    public decimal ID { get; set; }
}

Then the deserialization works without errors:

var testClassJson = "{\"Column0\": 12.0}";
var i = JsonConvert.DeserializeObject<TestClass>(testClassJson);

edited Feb 24, 2020 at 10:06

answered Feb 24, 2020 at 9:50

Tomas Chabada

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1 Comment

Yulia Koval Over a year ago

It works, but with type string [JsonProperty("Column0")] public string ID { get; set; }

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