英文:
Java convert a serialized object to another object
问题
我有两个Java数据对象，ObjectFloat用于反序列化传入的请求，ObjectInt用于传出的请求。我从HTTP调用中接收ObjectFloat，其中包含Float属性，对其进行反序列化，然后调用upgradeMembers()方法。因此，在将对象发送出去之前，我想要使用upgradeMembers()处理每个m_member1和m_member2属性，然后将其作为类型ObjectInt发送出去。
ObjectFloat:
public class ObjectFloat {
@SerializedName("member_number_1")
protected Float m_member1;
@SerializedName("member_number_2")
protected Float m_member2;
public void upgradeMembers() {
m_member1 = m_member1 * 100;
m_member2 = m_member2 * 100;
}
}
ObjectInt:
public class ObjectInt {
@SerializedName("member_number_1")
protected Integer m_member1;
@SerializedName("member_number_2")
protected Integer m_member2;
}
在将对象发送出去之前，您可以执行以下步骤将ObjectFloat转换为ObjectInt（不保留小数部分）：
- 使用
ObjectFloat对象调用upgradeMembers()方法，以确保m_member1和m_member2已被处理。 - 使用Gson库将
ObjectFloat对象转换为JSON字符串。 - 使用Gson库将JSON字符串转换回
ObjectInt对象。
以下是示例代码：
// 步骤1: 调用upgradeMembers()方法
ObjectFloat objectFloat = new ObjectFloat();
objectFloat.upgradeMembers();
// 步骤2: 将ObjectFloat转换为JSON字符串
Gson gson = new Gson();
String json = gson.toJson(objectFloat);
// 步骤3: 将JSON字符串转换为ObjectInt对象
ObjectInt objectInt = gson.fromJson(json, ObjectInt.class);
这样，您可以在不过度复杂化的情况下将ObjectFloat转换为ObjectInt并将其发送出去。
英文:
I have two java data objects ObjectFloat is for deseriealize incoming requests, and ObjectInt is for outgoing requests. I receive ObjectFloat from an HTTP call with Float attributes, deserialize it, and call upgradeMembers(). So before I send the object out I want to process each m_member1 and m_member1 attribute with upgradeMembers() and then send it out as a type ObjectInt.
Object float:
public class ObjectFloat {
@SerializedName("member_number_1")
protected Float m_member1;
@SerializedName("member_number_2")
protected Float m_member2;
public void upgradeMembers() {
m_member1 = m_member1 * 100
m_member1 = m_member1 * 100
}
}
Object integer:
public class ObjectInt {
@SerializedName("member_number_1")
protected Integer m_member1;
@SerializedName("member_number_2")
protected Integer m_member2;
}
How would I be able to covert ObjectFloat to ObjectInt before I send the object out? (I don't mind losing the decimals). I was thinking maybe I could call m_gson.toJson(objectfloat) and the serialize it back to with fromJson ObjectInt but when I send it out I would have to serialize it again with toJson() would that be overkill?
答案1
得分: 1
你可以添加一个以ObjectFloat为参数的构造函数。这样，你就可以基于ObjectFloat的实例创建ObjectInt的实例：
public class ObjectInt {
protected Integer m_member1;
protected Integer m_member2;
public ObjectInt() {
// 默认构造函数允许创建一个不依赖ObjectFloat的实例
}
public ObjectInt(ObjectFloat objectFloat) {
this.m_member1 = objectFloat.m_member1.intValue();
this.m_member2 = objectFloat.m_member2.intValue();
}
}
当你添加了这个构造函数后，你可以使用以下方式将ObjectFloat“转换”为ObjectInt：
ObjectInt objectInt = new ObjectInt(objectFloat);
英文:
You could add a constructor that takes ObjectFloat as a parameter. This way you can create an instance of ObjectInt based on an instance of ObjectFloat:
public class ObjectInt {
protected Integer m_member1;
protected Integer m_member2;
public ObjectInt() {
// default constructor allows creating an instance without ObjectFloat
}
public ObjectInt(ObjectFloat objectFloat) {
this.m_member1 = objectFloat.m_member1.intValue();
this.m_member2 = objectFloat.m_member2.intValue();
}
}
When you've added this constructor, you can "convert" an ObjectFloat into an ObjectInt with
ObjectInt objectInt = new ObjectInt(objectFloat)
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