JSON 決議為 Web API Post Call (Body)
{
"Name1": "Value1",
"Name2": "Value2",
"Name3": "Value3",
"Name4": "Value4"
}
讓我們稱這個物件為 NameValueObject
public class NameValueObject
{
public Name1 { get; set; }
public Name2 { get; set; }
public Name3 { get; set; }
public Name4 { get; set; }
}
我希望看到的是能夠決議 json 物件并將其轉換為串列
public async Task<List<NameValueObject>> EvaluateJson ([FromBody] NameValueObject request)
{
string serializerequest = JsonConvert.SerializeObject(request);
//What do i next , I just to want to get a list of the values sent through the json object
}
uj5u.com熱心網友回復:
你可以像這樣NameValueObject使用反射和Linq來獲取你的屬性的所有值:
var values = nameValueObject.GetType().GetProperties().Select(v => v.GetValue(nameValueObject));
如果您想同時獲得屬性的名稱和值,您可以創建一個這樣的字典:
var dictionary = nameValueObject.GetType().GetProperties().ToDictionary(v => v.Name, v => v.GetValue(nameValueObject));
uj5u.com熱心網友回復:
您可以創建 List< NameValueObject > 物件
var json= JsonConvert.Serialize(request);
List<NameValueObject> objects = JsonConvert.DeserializeObject<Dictionary<string, string>>(json)
.Select(i => new NameValueObject {Name=i.Key, Value=i.Value} )
.ToList();
或只是值串列
var json= JsonConvert.Serialize(request);
List<string> values=JsonConvert.DeserializeObject<Dictionary<string, string>>(json)
.Select(i => i.Value )
.ToList();
如果你想要 List< NameValueObject > ,你將不得不創建這個類
public class NameValueObject
{
public string Name { get; set; }
public string Value { get; set; }
}
或者我更喜歡這個
public async Task<List<NameValueObject>> EvaluateJson ([FromBody] JObject jsonObject)
{
return jsonObject.Properties()
.Select ( i => new NameValueObject {
Name = i.Name, Value = (string) i.Value }).ToList();
}
或只是值串列
List<string> Values = jsonObject.Properties().Select( i => (string) i.Value ).ToList();
轉載請註明出處,本文鏈接:https://www.uj5u.com/ruanti/379223.html
