C# 如何以编程方式从动态 JObject 获取属性
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How to get property from dynamic JObject programmatically
提问by dcdroid
I'm parsing a JSON string using the NewtonSoft JObject. How can I get values from a dynamic object programmatically? I want to simplify the code to not repeat myself for every object.
我正在使用 NewtonSoft JObject 解析 JSON 字符串。如何以编程方式从动态对象中获取值？我想简化代码，不要为每个对象重复我自己。
public ExampleObject GetExampleObject(string jsonString)
{
ExampleObject returnObject = new ExampleObject();
dynamic dynamicResult = JObject.Parse(jsonString);
if (!ReferenceEquals(dynamicResult.album, null))
{
//code block to extract to another method if possible
returnObject.Id = dynamicResult.album.id;
returnObject.Name = dynamicResult.album.name;
returnObject.Description = dynamicResult.albumsdescription;
//etc..
}
else if(!ReferenceEquals(dynamicResult.photo, null))
{
//duplicated here
returnObject.Id = dynamicResult.photo.id;
returnObject.Name = dynamicResult.photo.name;
returnObject.Description = dynamicResult.photo.description;
//etc..
}
else if..
//etc..
return returnObject;
}
Is there any way I can extract the code blocks in the "if" statements to a separate method e.g:
有什么方法可以将“if”语句中的代码块提取到单独的方法中，例如：
private void ExampleObject GetExampleObject([string of desired type goes here? album/photo/etc])
{
ExampleObject returnObject = new ExampleObject();
returnObject.Id = dynamicResult.[something goes here?].id;
returnObject.Name = dynamicResult.[something goes here?].name;
//etc..
return returnObject;
}
Is it even possible since we can't use reflection for dynamic objects. Or am I even using the JObject correctly?
甚至可能因为我们不能对动态对象使用反射。或者我什至正确使用 JObject 吗？
Thanks.
谢谢。
采纳答案by Connie Hilarides
Assuming you're using the Newtonsoft.Json.Linq.JObject, you don't need to use dynamic. The JObject class can take a string indexer, just like a dictionary:
假设您使用的是 Newtonsoft.Json.Linq.JObject，则不需要使用动态。JObject 类可以使用字符串索引器，就像字典一样：
JObject myResult = GetMyResult();
returnObject.Id = myResult["string here"]["id"];
Hope this helps!
希望这可以帮助！
回答by Hamit YILDIRIM
with dynamic keyword like below:
使用动态关键字如下：
private static JsonSerializerSettings jsonSettings;
private static T Deserialize<T>(string jsonData)
{
return JsonConvert.DeserializeObject<T>(jsonData, jsonSettings);
}
//if you know what will return
//如果你知道会返回什么
var jresponse = Deserialize<SearchedData>(testJsonString);
//if you know return object type you should sign it with json attributes like
//如果您知道返回对象类型，您应该使用json属性对其进行签名，例如
[JsonObject(MemberSerialization = MemberSerialization.OptIn)]
public class SearchedData
{
[JsonProperty(PropertyName = "Currency")]
public string Currency { get; set; }
[JsonProperty(PropertyName = "Routes")]
public List<List<Route>> Routes { get; set; }
}
// if you don't know the return type use dynamic as generic type
// 如果您不知道返回类型，请使用动态作为泛型类型
var jresponse = Deserialize<dynamic>(testJsonString);
回答by Ziad Akiki
Another way of targeting this is by using SelectToken(Assuming that you're using Newtonsoft.Json):
另一种针对此的方法是使用SelectToken（假设您正在使用Newtonsoft.Json）：
JObject json = GetResponse();
var name = json.SelectToken("items[0].name");
For a full documentation: https://www.newtonsoft.com/json/help/html/SelectToken.htm
完整文档：https: //www.newtonsoft.com/json/help/html/SelectToken.htm
