我正在尝试从一个“活动”发送客户类的对象,并在另一个“”中显示它。

客户类别的代码:

public class Customer {

    private String firstName, lastName, address;
    int age;

    public Customer(String fname, String lname, int age, String address) {

        firstName = fname;
        lastName = lname;
        age = age;
        address = address;
    }

    public String printValues() {

        String data = null;

        data = "First Name :" + firstName + " Last Name :" + lastName
        + " Age : " + age + " Address : " + address;

        return data;
    }
}

我想将其对象从一个“活动”发送到另一个“,然后在另一个活动”上显示数据。

我怎样才能做到这一点?


当前回答

正如在评论中提到的,这个答案打破了封装并紧密耦合了组件,这很可能不是您想要的。最好的解决方案可能是使对象可Parcelable或Serializable,正如其他响应所解释的那样。话虽如此,解决方案解决了问题。所以,如果你知道你在做什么:

使用带有静态字段的类:

public class Globals {
    public static Customer customer = new Customer();
}

在活动内部,您可以使用:

活动来源:

Globals.customer = myCustomerFromActivity;

活动目标:

myCustomerTo = Globals.customer;

这是传递活动信息的简单方法。

其他回答

在自定义类中创建两个方法,如下所示

public class Qabir {

    private int age;
    private String name;

    Qabir(){
    }

    Qabir(int age,String name){
        this.age=age; this.name=name;
    }   

    // method for sending object
    public String toJSON(){
        return "{age:" + age + ",name:\"" +name +"\"}";
    }

    // method for get back original object
    public void initilizeWithJSONString(String jsonString){

        JSONObject json;        
        try {
            json =new JSONObject(jsonString );
            age=json.getInt("age");
            name=json.getString("name");
        } catch (JSONException e) {
            e.printStackTrace();
        } 
    }
}

现在在您的发件人活动中这样做

Qabir q= new Qabir(22,"KQ");    
Intent in=new Intent(this,SubActivity.class);
in.putExtra("obj", q.toJSON());
startActivity( in);

在您的接收器中活动

Qabir q =new Qabir();
q.initilizeWithJSONString(getIntent().getStringExtra("obj"));
public class MyClass implements Serializable{
    Here is your instance variable
}

现在要在startActivity中传递该类的对象。只需使用此选项:

Bundle b = new Bundle();
b.putSerializable("name", myClassObject);
intent.putExtras(b);

这在这里有效,因为MyClass实现了Serializable。

创建类似bean类的类并实现Serializable接口。然后我们可以通过intent方法传递它,例如:

intent.putExtra("class", BeanClass);

然后从其他活动中获取,例如:

BeanClass cb = intent.getSerializableExtra("class");

我写了一个名为intentparser的库

它真的很容易使用将此添加到项目等级

allprojects {
        repositories {
            ...
            maven { url 'https://jitpack.io' }
        }
    }

将此添加到应用程序等级


dependencies {
            implementation 'com.github.lau1944:intentparser:v$currentVersion'
    }

使用扩展方法putObject传递对象

val testModel = TestModel(
            text = "hello world",
            isSuccess = false,
            testNum = 1,
            textModelSec = TextModelSec("second model")
)
startActivity(
     Intent(this, ActivityTest::class.java).apply {
          this.putObject(testModel)
     }
) 

从上一个活动获取对象


val testModel = intent.getObject(TestModel::class.java)

调用活动时

Intent intent = new Intent(fromClass.this,toClass.class).putExtra("myCustomerObj",customerObj);

在toClass.java中,通过

Customer customerObjInToClass = getIntent().getExtras().getParcelable("myCustomerObj");

请确保客户类实现parcelable

public class Customer implements Parcelable {

    private String firstName, lastName, address;
    int age;

    /* all your getter and setter methods */

    public Customer(Parcel in ) {
        readFromParcel( in );
    }

    public static final Parcelable.Creator CREATOR = new Parcelable.Creator() {
        public LeadData createFromParcel(Parcel in ) {
            return new Customer( in );
        }

        public Customer[] newArray(int size) {
            return new Customer[size];
        }
    };


    @Override
    public void writeToParcel(Parcel dest, int flags) {

        dest.writeString(firstName);
        dest.writeString(lastName);
        dest.writeString(address);
        dest.writeInt(age);
    }

    private void readFromParcel(Parcel in ) {

        firstName = in .readString();
        lastName  = in .readString();
        address   = in .readString();
        age       = in .readInt();
    }