java - 为什么一个对象在传入 Intent 后变为空?

标签 java android firebase firebase-realtime-database

点击客户姓名后,需要在谷歌地图上用标记标记他/她的位置。当我在弹出窗口上显示客户详细信息时,客户的纬度和经度通过 Intent 作为对象从弹出窗口类传递到 map Activity 类。在 map activity.java 中接收它时,它变为 null 并抛出空指针异常。

我的弹出窗口类代码(CustomerPupUp.java):

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_customer_pop_up);

    DisplayMetrics dm = new DisplayMetrics();
    getWindowManager().getDefaultDisplay().getMetrics(dm);

    int width = dm.widthPixels;
    int height = dm.heightPixels;

    getWindow().setLayout((int)(width),(int)(height*.7));

    listView = (ListView)findViewById(R.id.custListVw);
    database = FirebaseDatabase.getInstance();
    reference = database.getReference("Customers");

    list = new ArrayList<>();
    adapter = new ArrayAdapter<String>(this, R.layout.activity_cust_layout,R.id.cus, list);

    customer = new AddCustomer();
    reference.addValueEventListener(new ValueEventListener() {
        @Override
        public void onDataChange(DataSnapshot dataSnapshot) {
            for(DataSnapshot ds: dataSnapshot.getChildren() )
            {
                customer = ds.getValue(AddCustomer.class);
                Log.d("CustomerPopUp", "onDataChange: " + customer.getLat());
                list.add(customer.getFname().toString() + " " + customer.getLname().toString()  );
                customerList.add(customer);

            }
            listView.setAdapter(adapter);
            listView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
                @Override
                public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
                    Intent intent = new Intent(CustomerPopUp.this, MapsActivity.class);
                    intent.putExtra("CustomerData", customerList.get(position));
                    startActivity(intent);
                }
            });
        }

        @Override
        public void onCancelled(DatabaseError databaseError) {

        }
    });
}

以下代码段显示了我如何通过 MapsActivity.java 中的 Intent 获取对象

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_maps);

    try {
        database = FirebaseDatabase.getInstance();
        reference = database.getReference("Customers");

        customer= (AddCustomer) getIntent().getSerializableExtra("CustomerData");

        mMap.addMarker(new MarkerOptions().position(new LatLng(customer.getLat(), customer.getLng())).title(customer.getFname() + customer.getLname()));

    }catch (NullPointerException ex)
    {
        Log.d(TAG, "onCreate: " + ex.getMessage());
    }

    search = (EditText)findViewById(R.id.SearchText);
    gps =(ImageView) findViewById(R.id.GpsBtn);
    cust = (ImageView)findViewById(R.id.CusBtn);

    isServiceFine();

    getLocationPermission();

    cust.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            Intent intent = new Intent(MapsActivity.this, CustomerPopUp.class);
            startActivity(intent);
        }
    });
}

下面是AddCustomer类

public class AddCustomer implements Serializable {

private String fname;
private String lname;
private String email;
private String pass;
private String confPass;
private int num;
Double lat;
Double lng;


AddCustomer()
{

}

public AddCustomer(String FName, String LName, String EMail, String PWD, String CONFPWD, int NUM, Double latitude, Double longtitude)
{
    this.fname = FName;
    this.lname = LName;
    this.email = EMail;
    this.pass = PWD;
    this.confPass = CONFPWD;
    this.num = NUM;
    this.lat = latitude;
    this.lng = longtitude;

}

public String getFname() {
    return fname;
}

public void setFname(String fname) {
    this.fname = fname;
}

public String getLname() {
    return lname;
}

public void setLname(String lname) {
    this.lname = lname;
}

public String getEmail() {
    return email;
}

public void setEmail(String email) {
    this.email = email;
}

public String getPass() {
    return pass;
}

public void setPass(String pass) {
    this.pass = pass;
}

public String getConfPass() {
    return confPass;
}

public void setConfPass(String confPass) {
    this.confPass = confPass;
}

public int getNum() {
    return num;
}

public void setNum(int num) {
    this.num = num;
}

public Double getLat() {
    return lat;
}

public void setLat(Double lat) {
    this.lat = lat;
}

public Double getLng() {
    return lng;
}

public void setLng(Double lng) {
    this.lng = lng;
}

请帮我解决这个问题。

最佳答案

您只能通过Intent 传递ParcelableSerializable 对象。在这种情况下,对于您的自定义对象,您需要使其实现 ParcelableSerializable

我推荐使用 http://www.parcelabler.com/轻松生成您的代码。

参见:How can I make my custom objects Parcelable?

关于java - 为什么一个对象在传入 Intent 后变为空?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/50110098/

相关文章:

java - 是否可以在 Java 6 中为 RhinoScriptEngine 设置优化级别?

java - JLayeredPane 和 MouseAdapter 问题

java - 使用statement.executeQuery()执行两个不同的查询后使用ResultSet

android - 从android上的retrofit方法返回Callable

android - 使用 Apache POI 库解析出现在 Excel 工作表中的值

android - 运行 Android Junit 时找不到测试类异常

java - 如何从当前登录的用户 Firebase、Android 中检索值

java - 西蒙说计数器无法正常工作

firebase - 从 BigQuery 中删除带有 app_instance_id 的 firebase 实例

python - 如何在python中删除firebase电子邮件身份验证用户?