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Gson fromJson 拋出 ClassCastException

Gson API Reference

JsonParseException

  • gson.fromJson(jsonString, Map.class) 方法時拋出 JsonParseException 訊息
Caused by: com.google.gson.JsonParseException: The JsonDeserializer MapTypeAdapter failed to deserialize json object given the type interface java.util.Map

Gson Api Documentation

If the object that your are serializing/deserializing is a ParameterizedType (i.e. contains at least one type parameter and may be an array) then you must use the toJson(Object, Type) or fromJson(String, Type) method.
  • 因此在使用 gson.fromJson() 時第二個參數要帶入 TypeToken,也就是要告知 Gson 如何將 String 轉型。以下是範例 :

    1 2 3 4 5 6 7 8 9 10 11 12 13 14 import com.google.gson.Gson; import com.google.gson.GsonBuilder; import com.google.gson.reflect.TypeToken; public class MyDomainEntity<T>{ int age; String name; T vehicle; } Type type = new TypeToken< MyDomainEntity<Car>>(){}.getType(); MyDomainEntity map = gson.fromJson(jsonString, type);
  • 下面是 Pastgresql Jsonb 取出後直接將欄位值轉 String 的範例:

    1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 import com.google.gson.Gson; import com.google.gson.GsonBuilder; private Map<String, String> jsonbToMap(String jsonb) { Map<String, String> varMap = new LinkedHashMap<>(); if (jsonb != null) { Gson gson2 = new GsonBuilder().serializeNulls().create(); @SuppressWarnings("unchecked") Map<String, String> fromJson = (Map<String, String>) gson2 .fromJson(jsonb, Map.class); varMap.putAll(fromJson); } return varMap; }