c＃：DataSetへのDeserealization問題

Question

誰でもJSONの文字列解析をC＃で手助けできますか？

JsonConvert.DeserializeObject exception System.Runtime.Serialization.SerializationException: Member 'XmlSchema' was not found. 
09-17 07:41:27.075 I/mono-stdout(3691): JsonConvert.DeserializeObject exception System.Runtime.Serialization.SerializationException: Member 'XmlSchema' was not found. 
    at System.Runtime.Serialization.SerializationInfo.GetElement (System.String name, System.Type& foundType) [0x00020] in <3fd174ff54b146228c505f23cf75ce71>:0 
    at System.Runtime.Serialization.SerializationInfo.GetValue (System.String name, System.Type type) [0x0002e] in <3fd174ff54b146228c505f23cf75ce71>:0 
    at System.Data.DataSet.DeserializeDataSetSchema (System.Runtime.Serialization.SerializationInfo info, System.Runtime.Serialization.StreamingContext context, System.Data.SerializationFormat remotingFormat, System.Data.SchemaSerializationMode schemaSerializationMode) [0x000f5] in <eabff56c075e40a688404208276b3098>:0 
    at System.Data.DataSet.DeserializeDataSet (System.Runtime.Serialization.SerializationInfo info, System.Runtime.Serialization.StreamingContext context, System.Data.SerializationFormat remotingFormat, System.Data.SchemaSerializationMode schemaSerializationMode) [0x00000] in <eabff56c075e40a688404208276b3098>:0 
    at System.Data.DataSet..ctor (System.Runtime.Serialization.SerializationInfo inf 
o, System.Runtime.Serialization.StreamingContext context, System.Boolean ConstructSchema) [0x00069] in <eabff56c075e40a688404208276b3098>:0 
    at System.Data.DataSet..ctor (System.Runtime.Serialization.SerializationInfo info, System.Runtime.Serialization.StreamingContext context) [0x00000] in <eabff56c075e40a688404208276b3098>:0 
    at (wrapper dynamic-method) System.Object:lambda_method (System.Runtime.CompilerServices.Closure,object[]) 
    at Newtonsoft.Json.Serialization.JsonSerializerInternalReader.CreateISerializable (Newtonsoft.Json.JsonReader reader, Newtonsoft.Json.Serialization.JsonISerializableContract contract, Newtonsoft.Json.Serialization.JsonProperty member, System.String id) [0x0015b] in <c19705166c7c4a608e182e859c4de6d2>:0 
    at Newtonsoft.Json.Serialization.JsonSerializerInternalReader.CreateObject (Newtonsoft.Json.JsonReader reader, System.Type objectType, Newtonsoft.Json.Serialization.JsonContract contract, Newtonsoft.Json.Serialization.JsonProperty member, Newtonsoft.Json.Serialization.JsonCo 
ntainerContract containerContract, Newtonsoft.Json.Serialization.JsonProperty containerMember, System.Object existingValue) [0x00347] in <c19705166c7c4a608e182e859c4de6d2>:0 
    at Newtonsoft.Json.Serialization.JsonSerializerInternalReader.CreateValueInternal (Newtonsoft.Json.JsonReader reader, System.Type objectType, Newtonsoft.Json.Serialization.JsonContract contract, Newtonsoft.Json.Serialization.JsonProperty member, Newtonsoft.Json.Serialization.JsonContainerContract containerContract, Newtonsoft.Json.Serialization.JsonProperty containerMember, System.Object existingValue) [0x0006d] in <c19705166c7c4a608e182e859c4de6d2>:0 
    at Newtonsoft.Json.Serialization.JsonSerializerInternalReader.Deserialize (Newtonsoft.Json.JsonReader reader, System.Type objectType, System.Boolean checkAdditionalContent) [0x000db] in <c19705166c7c4a608e182e859c4de6d2>:0 
    at Newtonsoft.Json.JsonSerializer.DeserializeInternal (Newtonsoft.Json.JsonReader reader, System.Type objectType) [0x00053] in <c19705166c7c4a608e182e859c4de6d2>: 
0 
    at Newtonsoft.Json.JsonSerializer.Deserialize (Newtonsoft.Json.JsonReader reader, System.Type objectType) [0x00000] in <c19705166c7c4a608e182e859c4de6d2>:0 
    at Newtonsoft.Json.JsonConvert.DeserializeObject (System.String value, System.Type type, Newtonsoft.Json.JsonSerializerSettings settings) [0x0002d] in <c19705166c7c4a608e182e859c4de6d2>:0 
    at Newtonsoft.Json.JsonConvert.DeserializeObject[T] (System.String value, Newtonsoft.Json.JsonSerializerSettings settings) [0x00000] in <c19705166c7c4a608e182e859c4de6d2>:0 
    at Newtonsoft.Json.JsonConvert.DeserializeObject[T] (System.String value) [0x00000] in <c19705166c7c4a608e182e859c4de6d2>:0 

欠けているもの：私は、私は例外を受け取り、私はhttps://www.newtonsoft.com/json/help/html/DeserializeDataSet.htm

string json = @"{ 
      'Table1': [ 
        { 
        'id': 0, 
        'item': 'item 0' 
        }, 
        { 
        'id': 1, 
        'item': 'item 1' 
        } 
        ] 
       }"; 


    dataSet = JsonConvert.DeserializeObject<DataSet>(json); 

からサンプルを再現しよう....

この領域ではなく、新たなんですか？ XMLスキーマを追加する必要がありますか？事前に おかげ

Answer 1

はあなたが

**"Error: Parse error on line 1: 
'Table1': [{  'id':"** 

のような例外がありますが、あなたはそれを書く場合はそう

{ 
    "Table1": [{ 
     "id": 0 
    }, { 
     "id": 1 
    }] 
} 

代わりに「

はhttps://jsonlint.com であなたのJSONテキストを確認し、「使用する必要があります

よりもすべてが機能するはずです