将具有多个部分的对象从一个活动传递到另一个活动

Question

您好，我有一个要求，例如人员详细信息，因为我有多个详细信息，例如个人，教育，工作，例如在第一个活动中，我输入个人详细信息，然后是教育，然后是一个接一个的工作，我如何维护所有详细信息在单个对象中并通过应用程序请帮助我。

Answer 1

简单的解决方案：为此使用意图

---

Personal.class

public class Personal extends Activity {

    @Override 
    public void onCreate(Bundle savedInstanceState) {

        super.onCreate(savedInstanceState);
        setContentView(R.layout.personal);

        Intent i = new Intent(getApplicationContext(), Educational.class);
        i.putExtra("personal_details",<-get data from object->);
        startActivity(i);
    } 
}

教育类

public class Personal extends Activity {

    @Override 
    public void onCreate(Bundle savedInstanceState) {

        super.onCreate(savedInstanceState);
        setContentView(R.layout.personal);

        String personal_details; 

        Bundle extras = getIntent().getExtras();
        if (extras != null) {
        personal_details= extras.getString("personal_details");
        } 

        Intent i = new Intent(getApplicationContext(), educational.class);
        i.putExtra("personal_details",personal_details);
        i.putExtra("educational_details",<-get data from object->);
        startActivity(i);
    } 
}

Job.class

public class Personal extends Activity {

    @Override 
    public void onCreate(Bundle savedInstanceState) {

        super.onCreate(savedInstanceState);
        setContentView(R.layout.personal);

        String personal_details,educational_details; 

        Bundle extras = getIntent().getExtras();
        if (extras != null) {
        personal_details= extras.getString("personal_details");
        educational_details= extras.getString("educational_details");
        } 

        Intent i = new Intent(getApplicationContext(), FinalResult.class);
        i.putExtra("personal_details",personal_details);
        i.putExtra("educational_details",educational_details);
        i.putExtra("job_details",<-get data from object->);
        startActivity(i);
    } 
}

FinalResult.class

public class Personal extends Activity {

    @Override 
    public void onCreate(Bundle savedInstanceState) {

        super.onCreate(savedInstanceState);
        setContentView(R.layout.personal);
        String personal_details,educational_details,job_details; 

        Bundle extras = getIntent().getExtras();
        if (extras != null) {
        personal_details= extras.getString("personal_details");
        educational_details= extras.getString("educational_details");
        job_details= extras.getString("job_details");
        } 
    } 
}

---

{EDIT-1}

• 看看我在different storage options in android - Click Here 中的回答之一
• 您可以使用application variable 或shared preferences 来实现它，但我不推荐它。 ！
• 尝试将其存储在 POJO（普通旧 Java 类）中

{EDIT-2}

嗯，如果我理解正确您的问题:: 您已连接到互联网（Wifi、有线等），基本上您是在尝试在没有网络连接时显示一个对话框！ .... 可以求助Broadcast receivers...

试试这个:: 设置广播接收器以在没有网络连接时触发 Intent .... 编写代码以捕获该意图并弹出dialog ...。在此对话框中，为用户提供重新连接连接的选项！

Answer 2

请注意，您需要在Intent 中捆绑一个对象，该对象必须实现Parcelable 接口。这是取自Parcelable documentation的一个常见实现...

public class MyParcelable implements Parcelable {
     private int mData;

     public int describeContents() {
         return 0;
     }

     public void writeToParcel(Parcel out, int flags) {
         out.writeInt(mData);
     }

     public static final Parcelable.Creator<MyParcelable> CREATOR
             = new Parcelable.Creator<MyParcelable>() {
         public MyParcelable createFromParcel(Parcel in) {
             return new MyParcelable(in);
         }

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

     private MyParcelable(Parcel in) {
         mData = in.readInt();
     }
 }

完成此操作后，只需在 Intent... 中传递对象...

Intent i = new Intent(getApplicationContext(), ACTIVITY_TO_START.class);
i.putExtra("extra_key", new MyParcelable());
startActivity(i);

检索开始活动的对象...

Bundle extras = getIntent().getExtras();

if(extras != null && extras.containsKey("extra_key"))
{
    MyParcelable p = (MyParecelable)extras.getParcelable("extra_key");
}