Question

我正在使用Gson将JSON转换为Java对象。我在json中有一个字段，我们明智的服务人员编码为数组或对象，具体取决于数据库中有多少项返回。问题是如何模拟Java对象传递到Gson转换器，以便我可以处理这两种类型？

json = new Gson().fromJson(reader, Order.class);

Java Classes只能正确解析JSON数组

public class Order {
    private Detail detail
}

public class Detail {
    public String id;
    public List<Type> types;
    //// getters and setters
}

public class Type {
    public String typeId;
    public String typeName
    //// getters and setters
}

JSON数据数组

{
    "detail":{
        "id":"1234565",
        "types":{
            "type":[
                {"typeId":"1246565","typeName":"MyName1"},
                {"typeId":"1444445","typeName":"MyName2"}
            ]
        }
    }
}

JSON数据对象

{
    "detail":{
        "id":"1234565",
        "types":{
            "type":{"typeId":"1246565","typeName":"MyName1"}
        }
    }
}

Answer 1

我想通过使用通用对象来解决这个问题。所以现在我仍然可以使用GSON进行调用，而且只有当我需要重构对象时才需要进行那么多的手动解析，而且只需要那个特定的对象。

json = new Gson().fromJson(reader, Order.class);

主要对象

public class Order {
    private Detail detail
}

详情课程

public class Detail {
    public String id;
    public Types types;

    //// getters and setters
    public Types getTypes() {
        return types;
    }

    public void setTypes(Types types) {
        this.types = types;
    }
}

包含一个名为type的通用Object的Types类，因此它可以存储List 或者是LinkedTreeMap，它是JSONObject被解析的方式。然后，您必须手动将其插入到新的Type对象中。

public class Types {
    private Object type;

    //// getters and setters

    public Object getType() {
        return type;
    }

    public void setType(Object type) {
        this.type = type;
    }

    public List<Type> getListType() {
        if (type instanceof List) {
            return (List<Type>) type;
        } else
            return null;
    }

    public Type getObjectType() {
        if (type instanceof Type) {
            return (Type) type;
        } else if (type instanceof Map) {
            Map<String, String> map = (Map<String, String>)type;
            Type newType = new Type((String)map.get("typeId"), (String)map.get("typeName"));
            return newType;
        }
        return null;
    }

public class Type {
    private String typeId;
    private String typeName;

    public Type() {}

    public Type(String id, String name) {
        this.typeId = id;
        this.typeName = name;
    }

    //// getters and setters
    public String getTypeId() {
        return typeId;
    }

    public void setTypeId(String typeId) {
        this.typeId = typeId;
    }

    public String getTypeName() {
        return typeName;
    }

    public void setTypeName(String typeName) {
        this.typeName = typeName;
    }
}

Answer 2

嗟!!

JsonArray typeArray; 
JsonElement typeElement = types.get("type"); 
if (typeElement.isJsonObject()) { 
    typeArray = new JsonArray();
    typeArray.add(typeElement); 
}
else {
    typeArray = (JsonArray)typeElement;
}
for (int i = 0; i < typeArray.size(); i++) {
    JsonObject typeObject = (JsonObject)(typeArray.get(i));
    typeObject.doSomethingWithThis();
}

如何使用GSON WHen对象解析JSON是JSONObject还是JSONArray

2 个答案: