Json Serialize Ignore Property

Question

Hi, I want to delete a property in my model. When I serialize json, I want json result not to appear. how is this possible. Class

    public class SkuVariant
    {

        public int fldVariant1TypeId { get; set; }

        public string fldVariant1TypeName { get; set; }

        public int fldVariant2TypeId { get; set; }
        public string fldVariant2TypeName { get; set; }

        public int fldVariant3TypeId { get; set; }
        public string fldVariant3TypeName { get; set; }

        [JsonIgnore(Condition = JsonIgnoreCondition.Always)]
        public string? Sku { get; set; }
    
    }

Dapper Query

 Select AL.[Articel.Code] AS Sku , 168 AS fldVariant1TypeId, BD.MainColor_Turkish AS fldVariant1TypeName, 
				169 AS fldVariant2TypeId , BS.Value AS fldVariant2TypeName ,
				169 AS fldVariant3TypeId, BSC.Value AS fldVariant3TypeName  
                 from vw_ArticelHorVerValueInfo AL

Controller Json

  var jsons = JsonConvert.SerializeObject(first);

Answer 1

Hi @Ali Yılmaz, If you want to ignore specified property when serializing, You can use [JsonIgnore] under System.Text.Json.Serialization namespace. Please refer to this simple demo:

using System.Text.Json.Serialization;

namespace WebApplication1.Model
{
    public class TestModel
    {
        public int Id { get; set; }
        public string Name { get; set; }
        [JsonIgnore]
        public string Country { get; set; }
    }
}

Controller

 [HttpGet]

    public IActionResult Get()
    {

        List<TestModel> models = new List<TestModel>();
        TestModel model = new TestModel()
        {
            Id = 1,
            Name = "aaaa",
            Country = "USA"
        };

        TestModel model2 = new TestModel()
        {
            Id = 2,
            Name = "bbb",
            Country = "CN"
        };

        models.Add(model);
        models.Add(model2);
        var json = JsonSerializer.Serialize(models);
        return Ok(json);
    }

Result

You can find it ignore Country property successfully.

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