为每个联系人获取相同的显示

时间:2013-09-22 09:23:30

标签: android contacts

我正在使用虚拟数据创建一个联系人管理器,我目前有一个包含联系人ArrayList的公共字段,即public static ArrayList contactList = new ArrayList();

点击特定联系人后(从查看联系人屏幕),我将进入编辑联系人屏幕,其中显示该特定联系人的所有字段。即名称,手机,地址,电子邮件等。列表中的第一个联系人在单击时会正确显示其详细信息。但是,当我返回到视图联系人屏幕时,它仍然显示第一个联系人的字段。

在我的MainActivity.java类中:

        listView.setOnItemClickListener(new AdapterView.OnItemClickListener(){
        @Override
            public void onItemClick(AdapterView<?> parentView, View clickedView, int clickedViewPosition, long id){
                TextView firstName = (TextView) findViewById(R.id.firstName);



                String fName = firstName.getText().toString();  


                Intent theIntent = new Intent(getApplication(), EditContact.class);
                theIntent.putExtra("firstName", fName);
                startActivity(theIntent);

            }

        });

我将联系人的firstName传递给putExtra(),这将我带到EditContact类,我在其中运行ArrayList并获取具有相同firstName的联系人,然后获取其所有字段(存储在相应的文本中)框)。

@Override
public void onCreate(Bundle savedInstanceState) {

    // Get saved data if there is any

    super.onCreate(savedInstanceState);

    // Designate that add_new_contact.xml is the interface used

    setContentView(R.layout.edit_contact);

    // Initialize the EditText objects

    firstName = (EditText) findViewById(R.id.firstName);
    lastName = (EditText) findViewById(R.id.lastName);
    dateOfBirth = (EditText) findViewById(R.id.dateOfBirth);
    mobileNumber = (EditText) findViewById(R.id.mobileNumber);
    homeNumber = (EditText) findViewById(R.id.homeNumber);
    workNumber = (EditText) findViewById(R.id.workNumber);
    homeEmail = (EditText) findViewById(R.id.emailAddressHome);
    workEmail = (EditText) findViewById(R.id.emailAddressWork);
    homeAddress = (EditText) findViewById(R.id.homeAddress);
    workAddress = (EditText) findViewById(R.id.workAddress);

    Intent theIntent = getIntent();

    String thefirstName = theIntent.getStringExtra("firstName");

    for (Contact con: MainActivity.contactList){

        if (con.getFirstName().equals(thefirstName)){
            firstName.setText(con.getFirstName());
            lastName.setText(con.getLastName());
            dateOfBirth.setText(con.getDateOfBirth());
            mobileNumber.setText(con.getMobileNumber());
            homeNumber.setText(con.getHomeNumber());
            workNumber.setText(con.getWorkNumber());
            homeEmail.setText(con.getHomeEmail());
            workEmail.setText(con.getWorkEmail());
            homeAddress.setText(con.getHomeAddress());
            workAddress.setText(con.getWorkAddress());

        }

    }


}

当我点击其他联系人时,有谁知道为什么putExtra()中传递的firstName不会改变?

0 个答案:

没有答案