我正在使用 Newtonsoft JSON 序列化程序，它适用于大多数对象。

不幸的是，当我尝试序列化一个大对象时，我得到了一个 JsonSerializationException，其中一个成员抛出了一个 NullReferenceException。

有没有办法忽略有问题的成员并序列化对象的其余部分？

我在想也许在 JsonSerializerSettings 中？

这是我想要做的简化版本:

private class TestExceptionThrowingClass
{
    public string Name { get { return "The Name"; } }
    public string Address { get { throw new NullReferenceException(); } }
    public int Age { get { return 30; } }
}

[Test]
public void CanSerializeAClassWithAnExceptionThrowingMember()
{ 
    // Arrange
    var toSerialize = new TestExceptionThrowingClass();

    // Act

    var serializerSettings = new Newtonsoft.Json.JsonSerializerSettings();
    serializerSettings.MaxDepth = 5;
    serializerSettings.ReferenceLoopHandling = Newtonsoft.Json.ReferenceLoopHandling.Ignore;
    serializerSettings.MissingMemberHandling = Newtonsoft.Json.MissingMemberHandling.Ignore;
    serializerSettings.NullValueHandling = Newtonsoft.Json.NullValueHandling.Ignore;
    serializerSettings.ObjectCreationHandling = Newtonsoft.Json.ObjectCreationHandling.Reuse;
    serializerSettings.DefaultValueHandling = Newtonsoft.Json.DefaultValueHandling.Ignore;

    var result = Newtonsoft.Json.JsonConvert.SerializeObject(toSerialize);

    // Assert
    Assert.That(result, Is.EqualTo(@"{""Name"":""The Name"",""Age"":30}"));
}

这是堆栈跟踪:

at Newtonsoft.Json.Serialization.DynamicValueProvider.GetValue(Object target) 
at Newtonsoft.Json.Serialization.JsonSerializerInternalWriter.CalculatePropertyValues(JsonWriter writer, Object value, JsonContainerContract contract, JsonProperty member, JsonProperty property, JsonContract& memberContract, Object& memberValue) 
at Newtonsoft.Json.Serialization.JsonSerializerInternalWriter.SerializeObject(JsonWriter writer, Object value, JsonObjectContract contract, JsonProperty member, JsonContainerContract collectionContract, JsonProperty containerProperty) 
at Newtonsoft.Json.Serialization.JsonSerializerInternalWriter.SerializeValue(JsonWriter writer, Object value, JsonContract valueContract, JsonProperty member, JsonContainerContract containerContract, JsonProperty containerProperty) 
at Newtonsoft.Json.Serialization.JsonSerializerInternalWriter.Serialize(JsonWriter jsonWriter, Object value, Type objectType) 
at Newtonsoft.Json.JsonSerializer.SerializeInternal(JsonWriter jsonWriter, Object value, Type objectType) 
at Newtonsoft.Json.JsonConvert.SerializeObject(Object value, Type type, Formatting formatting, JsonSerializerSettings settings) 
at Newtonsoft.Json.JsonConvert.SerializeObject(Object value) 
at AspectsProject.Aspects.CachingPolicy.CachingPolicyCacheKeyCreatorTests.CanSerializeAClassWithAnExceptionThrowingMember() in D:\Dev\test.cs:line 169
    --NullReferenceException 
at AspectsProject.Aspects.CachingPolicy.CachingPolicyCacheKeyCreatorTests.TestExceptionThrowingClass.get_Address() in D:\Dev\test.cs:line 149 
at GetAddress(Object ) 
at Newtonsoft.Json.Serialization.DynamicValueProvider.GetValue(Object target)

如果有人知道会这样做，我很乐意使用不同的 JSON 序列化程序。

最佳答案

忽略错误的更简单方法:

JsonSerializerSettings settings = new JsonSerializerSettings();
settings.Error = (serializer,err) => {
    err.ErrorContext.Handled = true;
}

或

settings.Error = (serializer,err) => err.ErrorContext.Handled = true;

关于c# - 忽略序列化为 JSON 时抛出异常的类成员，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20962316/