System.Text.Json序列化dynamic对象时无法忽略null值的问题

我需要将dynamic对象（实际场景是SQL查询结果存入dynamic对象，包含大量null列）序列化为JSON时，抑制所有null属性。但使用System.Text.Json.JsonSerializer.Serialize()时，不管设置哪种JsonIgnoreCondition选项都无效——这个选项对强类型有效，但对dynamic对象（比如ExpandoObject）不起作用。

复现代码 #

using System.Dynamic;
using System.Text.Json.Serialization;
using System.Text.Json;

var foo = new Foo();
var fooJson = JsonSerializer.Serialize(foo, new JsonSerializerOptions() { WriteIndented = true});
var fooJson1 = JsonSerializer.Serialize(foo, new JsonSerializerOptions() { WriteIndented = true, DefaultIgnoreCondition = JsonIgnoreCondition.WhenWritingDefault});
var fooJson2 = JsonSerializer.Serialize(foo, new JsonSerializerOptions() {WriteIndented = true, DefaultIgnoreCondition = JsonIgnoreCondition.WhenWritingNull});

Console.WriteLine("fooJson:");
Console.WriteLine(fooJson);
Console.WriteLine("fooJson1:");
Console.WriteLine(fooJson1);
Console.WriteLine("fooJson2:");
Console.WriteLine(fooJson2);

dynamic dynamicFoo = new ExpandoObject();
dynamicFoo.Prop1 = "something";
dynamicFoo.Prop2 = (int?)null;
dynamicFoo.Prop3 = (string)null;

var dynamicFooJson = JsonSerializer.Serialize(dynamicFoo, new JsonSerializerOptions() { WriteIndented = true});
var dynamicFooJson1 = JsonSerializer.Serialize(dynamicFoo, new JsonSerializerOptions() { WriteIndented = true, DefaultIgnoreCondition = JsonIgnoreCondition.WhenWritingDefault});
var dynamicFooJson2 = JsonSerializer.Serialize(dynamicFoo, new JsonSerializerOptions() {WriteIndented = true, DefaultIgnoreCondition = JsonIgnoreCondition.WhenWritingNull});

Console.WriteLine("dynamicFooJson:");
Console.WriteLine(dynamicFooJson);
Console.WriteLine("dynamicFooJson1:");
Console.WriteLine(dynamicFooJson1);
Console.WriteLine("dynamicFooJson2:");
Console.WriteLine(dynamicFooJson2);


public class Foo
{
    public string Prop1 {get;set;}
    public int Prop2 {get;set;}
    public int? Prop3 {get;set;}
}

执行输出 #

fooJson:
{
  "Prop1": null,
  "Prop2": 0,
  "Prop3": null
}
fooJson1:
{}
fooJson2:
{
  "Prop2": 0
}
dynamicFooJson:
{
  "Prop1": "something",
  "Prop2": null,
  "Prop3": null
}
dynamicFooJson1:
{
  "Prop1": "something",
  "Prop2": null,
  "Prop3": null
}
dynamicFooJson2:
{
  "Prop1": "something",
  "Prop2": null,
  "Prop3": null
}

解决方案 #

方法1：手动过滤Null值 #

ExpandoObject实现了IDictionary<string, object>接口，可先转为字典过滤掉值为null的项，再序列化：

dynamic dynamicFoo = new ExpandoObject();
dynamicFoo.Prop1 = "something";
dynamicFoo.Prop2 = (int?)null;
dynamicFoo.Prop3 = (string)null;

// 过滤null值
var filteredDict = ((IDictionary<string, object>)dynamicFoo)
    .Where(kv => kv.Value != null)
    .ToDictionary(kv => kv.Key, kv => kv.Value);

var json = JsonSerializer.Serialize(filteredDict, new JsonSerializerOptions { WriteIndented = true });
Console.WriteLine(json);

输出：

{
  "Prop1": "something"
}

方法2：自定义全局JsonConverter #

如果需要统一处理所有ExpandoObject的序列化，可自定义转换器跳过null值：

public class ExpandoObjectNullIgnoringConverter : JsonConverter<ExpandoObject>
{
    public override ExpandoObject Read(ref Utf8JsonReader reader, Type typeToConvert, JsonSerializerOptions options)
    {
        var dict = JsonSerializer.Deserialize<Dictionary<string, object>>(ref reader, options);
        var expando = new ExpandoObject() as IDictionary<string, object>;
        if (dict != null)
        {
            foreach (var kv in dict)
            {
                expando.Add(kv);
            }
        }
        return (ExpandoObject)expando;
    }

    public override void Write(Utf8JsonWriter writer, ExpandoObject value, JsonSerializerOptions options)
    {
        writer.WriteStartObject();
        foreach (var kv in value as IDictionary<string, object>)
        {
            if (kv.Value != null)
            {
                writer.WritePropertyName(kv.Key);
                JsonSerializer.Serialize(writer, kv.Value, options);
            }
        }
        writer.WriteEndObject();
    }
}

使用方式：

var options = new JsonSerializerOptions
{
    WriteIndented = true,
    Converters = { new ExpandoObjectNullIgnoringConverter() }
};

var json = JsonSerializer.Serialize(dynamicFoo, options);
Console.WriteLine(json);

内容的提问来源于stack exchange，提问作者Mr. T