我有一个包含以下字段的课程。当需要调用外部rest API方法时,这些属性用于序列化为json对象。
public class Customer
{
[JsonProperty(PropertyName = "email")]
public string Email { get; set; }
[JsonProperty(PropertyName = "prop[listId]")]
public string Test{ get; set; }
// there are lot of properties
}
在属性名称Test
中,外部API服务调用需要遵循json文件名格式。
prop[7]
在我的情况下,这个7
可以根据test,dev和prod等环境进行更改。所以我正在寻找一种方法将listId值移动到app.config中。
我试图按照以下方式执行此操作,但不允许这样做。对于listIdValue
,如果指定常量值,它将起作用。
private string listIdValue = ConfigurationManager.AppSettings["ListIdValue"];
[JsonProperty(PropertyName = "prop["+listIdValue +"]")]
public string Test{ get; set; }
答案 0 :(得分:9)
您必须覆盖DefaultContractResolver
并实施自己的机制来提供PropertyName
(使用JSON)。我将提供一个完整的示例代码,以显示运行时生成的PropertyName
的反序列化和序列化。目前,它会将Test
字段修改为Test5
(在所有模型中)。您应该实现自己的机制(使用属性,保留名称,表或其他。
class Program
{
static void Main(string[] args)
{
var customer = new Customer() {Email = "asd@asd.com", Test = "asdasd"};
var a = Serialize(customer, false);
var b = Serialize(customer, true);
Console.WriteLine(a);
Console.WriteLine(b);
var desA = Deserialize<Customer>(a, false);
var desB = Deserialize<Customer>(b, true);
Console.WriteLine("TestA: {0}", desA.Test);
Console.WriteLine("TestB: {0}", desB.Test);
}
static string Serialize(object obj, bool newNames)
{
JsonSerializerSettings settings = new JsonSerializerSettings();
settings.Formatting = Formatting.Indented;
if (newNames)
{
settings.ContractResolver = new CustomNamesContractResolver();
}
return JsonConvert.SerializeObject(obj, settings);
}
static T Deserialize<T>(string text, bool newNames)
{
JsonSerializerSettings settings = new JsonSerializerSettings();
settings.Formatting = Formatting.Indented;
if (newNames)
{
settings.ContractResolver = new CustomNamesContractResolver();
}
return JsonConvert.DeserializeObject<T>(text, settings);
}
}
class CustomNamesContractResolver : DefaultContractResolver
{
protected override IList<JsonProperty> CreateProperties(System.Type type, MemberSerialization memberSerialization)
{
// Let the base class create all the JsonProperties
// using the short names
IList<JsonProperty> list = base.CreateProperties(type, memberSerialization);
// Now inspect each property and replace the
// short name with the real property name
foreach (JsonProperty prop in list)
{
if (prop.UnderlyingName == "Test") //change this to your implementation!
prop.PropertyName = "Test" + 5;
}
return list;
}
}
public class Customer
{
[JsonProperty(PropertyName = "email")]
public string Email { get; set; }
public string Test { get; set; }
}
输出:
{
"email": "asd@asd.com",
"Test": "asdasd"
}
{
"email": "asd@asd.com",
"Test5": "asdasd"
}
TestA: asdasd
TestB: asdasd
如您所见,当我们使用Serialize(..., false)
时,字段的名称为Test
,当我们使用Serialize(..., true)
时,字段的名称为Test5
,正如所料。这也适用于反序列化。
我已将此答案用作检查我的答案:https://stackoverflow.com/a/20639697/773879
答案 1 :(得分:1)
定义不同的配置模式,如Debug / Release / QA / Staging
然后为每一个添加编译符号。并在您的代码中执行以下操作:
以下我假设你定义了:QA和STAGING
public class Customer
{
[JsonProperty(PropertyName = "email")]
public string Email { get; set; }
#if QA
[JsonProperty(PropertyName = "prop[QA_ID]")]
#elif STAGING
[JsonProperty(PropertyName = "prop[STAGING_ID]")]
#endif
public string Test{ get; set; }
// there are lot of properties
}
您也可以将这些配置用于自动部署,这将节省您的时间。