我正在尝试使用Json.NET库反序列化JSON。我收到的JSON如下:
{
"responseHeader": {
"zkConnected": true,
"status": 0,
"QTime": 2
},
"suggest": {
"mySuggester": {
"Ext": {
"numFound": 10,
"suggestions": [
{
"term": "Extra Community",
"weight": 127,
"payload": ""
},
{
"term": "External Video block",
"weight": 40,
"payload": ""
},
{
"term": "Migrate Extra",
"weight": 9,
"payload": ""
}
]
}
}
}
}
问题在于,您可以在其中看到的“ Ext” 是查询字符串中传递的参数的一部分,并且总是不同的。我只想获取分配给术语“ term”的值。
我尝试了类似的方法,但不幸的是,它不起作用:
public class AutocompleteResultsInfo
{
public AutocompleteResultsInfo()
{
this.Suggest = new Suggest();
}
[JsonProperty("suggest")]
public Suggest Suggest { get; set; }
}
public class Suggest
{
[JsonProperty("mySuggester")]
public MySuggesterElement MySuggesterElement { get; set; }
}
public struct MySuggesterElement
{
public MySuggester MySuggester;
public string JsonString;
public static implicit operator MySuggesterElement(MySuggester MySuggester) =>new MySuggesterElement { MySuggester = MySuggester };
public static implicit operator MySuggesterElement(string String) => new MySuggesterElement { JsonString = String };
}
public class MySuggester
{
[JsonProperty("suggestions")]
public Suggestions[] Suggestions { get; set; }
}
public class Suggestions
{
[JsonProperty("term")]
public string Autocopmplete { get; set; }
}
internal class SuggesterElementConverter : JsonConverter
{
public override bool CanConvert(Type t)
{
return t == typeof(MySuggesterElement) || t == typeof(MySuggesterElement?);
}
public override object ReadJson(JsonReader reader, Type objectType, object existingValue, JsonSerializer serializer)
{
switch (reader.TokenType)
{
case JsonToken.String:
case JsonToken.Date:
var stringValue = serializer.Deserialize<string>(reader);
return new MySuggesterElement { JsonString = stringValue };
case JsonToken.StartObject:
var objectValue = serializer.Deserialize<MySuggester>(reader);
return new MySuggesterElement { MySuggester = objectValue };
}
throw new Exception("Cannot unmarshal type MySuggesterElement");
}
public override void WriteJson(JsonWriter writer, object untypedValue, JsonSerializer serializer)
{
var value = (MySuggesterElement)untypedValue;
if (value.JsonString != null)
{
serializer.Serialize(writer, value.JsonString);
return;
}
if (value.MySuggester != null)
{
serializer.Serialize(writer, value.MySuggester);
return;
}
throw new Exception("Cannot marshal type CollationElements");
}
public static readonly SuggesterElementConverter Singleton = new SuggesterElementConverter();
}
public class AutocompleteConverter
{
public static readonly JsonSerializerSettings Settings = new JsonSerializerSettings
{
MetadataPropertyHandling = MetadataPropertyHandling.Ignore,
DateParseHandling = DateParseHandling.None,
Converters =
{
SuggesterElementConverter.Singleton
}
};
}
var results = JsonConvert.DeserializeObject<AutocompleteResultsInfo>(resultJson, AutocompleteConverter.Settings);
非常感谢您的帮助。
亲切的问候, Wojciech
答案 0 :(得分:3)
您可以将“ mySuggester”解码为字典: 公共课建议
public class Suggest
{
[JsonProperty("mySuggester")]
public Dictionary<string, MySuggester> MySuggester { get; set; }
}
然后,您将可以使用查询字符串参数访问建议:
var variablePropertyName = "Ext";
var result = JsonConvert.DeserializeObject<AutocompleteResultsInfo>(_json);
var suggestions = result.Suggest.MySuggester[variablePropertyName].Suggestions;
如果您不知道属性名称,也可以在字典中查找它:
var variablePropertyName = result.Suggest.MySuggester.Keys.First();
答案 1 :(得分:1)
如果您不需要反序列化整个json字符串,则可以使用JsonTextReader。示例:
private static IEnumerable<string> GetTerms(string json)
{
using (JsonTextReader reader = new JsonTextReader(new StringReader(json)))
{
while (reader.Read())
{
if (reader.TokenType == JsonToken.PropertyName && reader.Value.Equals("term"))
{
string term = reader.ReadAsString();
yield return term;
}
}
}
}
使用代码:
string json = @"{
""responseHeader"": {
""zkConnected"": true,
""status"": 0,
""QTime"": 2
},
""suggest"": {
""mySuggester"": {
""Ext"": {
""numFound"": 10,
""suggestions"": [
{
""term"": ""Extra Community"",
""weight"": 127,
""payload"": """"
},
{
""term"": ""External Video block"",
""weight"": 40,
""payload"": """"
},
{
""term"": ""Migrate Extra"",
""weight"": 9,
""payload"": """"
}
]
}
}
}
}";
IEnumerable<string> terms = GetTerms(json);
foreach (string term in terms)
{
Console.WriteLine(term);
}
答案 2 :(得分:1)
如果您只需要包含术语的对象,而无需其他任何内容,
您可以通过使用JSON.Net中的JObject
接口直接使用JSON值。
var parsed = JObject.Parse(jsonString);
var usingLinq = (parsed["suggest"]["mySuggester"] as JObject)
.Descendants()
.OfType<JObject>()
.Where(x => x.ContainsKey("term"));
var usingJsonPath = parsed.SelectTokens("$.suggest.mySuggester.*.*[?(@.term)]")
.Cast<JObject>();