是否可以让 JObject.Parse 忽略缺失的字段?
从下面的示例中,您可以看到我已经声明了一个类Address
并JsonProperty
用于指定备用字段名称。
我提供了 3 个示例,其中 3 个 JSON 字符串的结构略有不同,只有示例 1 匹配并返回一个对象,示例 2 和示例 3 返回 null,因为缺少字段。
如果没有提供,有没有办法使用其他 JsonProperty 允许它们被忽略?
public class Address
{
[JsonProperty("flat_number")]
public string FlatNumber { get; set; }
[JsonProperty("house_number")]
public string HouseNumber { get; set; }
[JsonProperty("address")]
public string Address1 { get; set; }
[JsonProperty("address2")]
public string Address2 { get; set; }
[JsonProperty("town")]
public string Town { get; set; }
[JsonProperty("postcode")]
public string Postcode { get; set; }
}
private static T TryParse<T>(string json) where T : new()
{
var jSchemaGenerator = new JSchemaGenerator();
const string license = "license";
License.RegisterLicense(license);
var jSchema = jSchemaGenerator.Generate(typeof(T));
var jObject = JObject.Parse(json);
return jObject.IsValid(jSchema) ? JsonConvert.DeserializeObject<T>(json) : default(T);
}
//Example 1 with house_number and flat_number
const string json = "{\"house_number\":\"40\",\"flat_number\":\"82\",\"address\":\"Somewhere\",\"address2\":\"Over\",\"town\":\"There\",\"postcode\":\"ZZ991AA\"}";
//Example 2 with house_number but not flat_number
//const string json = "{\"house_number\":\"40\",\"address\":\"Somewhere\",\"address2\":\"Over\",\"town\":\"There\",\"postcode\":\"ZZ991AA\"}";
//Example 3 with flat_number but not house_number
//const string json = "{\"flat_number\":\"82\",\"address\":\"Somewhere\",\"address2\":\"Over\",\"town\":\"There\",\"postcode\":\"ZZ991AA\"}";
var tryParse = TryParse<AddressTest>(json);
if (tryParse != null)
{
}