2010-06-11 195 views
0

首先,我很抱歉我的英語...從android手機獲取電話號碼

我有一個問題,從聯繫人獲取電話號碼。

這是我的代碼

import android.app.ListActivity; 
import android.database.Cursor; 
import android.os.Bundle; 
import android.provider.ContactsContract; 
import android.widget.SimpleAdapter; 
import android.widget.Toast; 
import java.util.ArrayList; 
import java.util.HashMap; 


public class TestContacts extends ListActivity { 


private ArrayList<HashMap<String,String>> list = new ArrayList<HashMap<String,String>>(); 

private SimpleAdapter numbers; 

/** Called when the activity is first created. */ 
@Override 
public void onCreate(Bundle savedInstanceState) { 
super.onCreate(savedInstanceState); 
    setContentView(R.layout.contacts); 
    numbers = new SimpleAdapter( 
    this, 
list, 
R.layout.main_item_two_line_row, 
new String[] { "line1","line2" }, 
new int[] { R.id.text1, R.id.text2 } ); 
    setListAdapter(numbers); 



    Cursor cursor = getContentResolver().query(ContactsContract.Contacts.CONTENT_URI, 
    null, null, null, null); 
    while (cursor.moveToNext()) { 
    String contactId = cursor.getString(cursor.getColumnIndex( 
    ContactsContract.Contacts._ID)); 
    String hasPhone = cursor.getString(cursor.getColumnIndex( 
    ContactsContract.Contacts.HAS_PHONE_NUMBER)); 

        //check if the contact has a phone number 
    if (Boolean.parseBoolean(hasPhone)) { 

Cursor phones = getContentResolver().query( 
    ContactsContract.CommonDataKinds.Phone.CONTENT_URI, 
    null, 
    ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = "+ contactId, 
    null, null); 
while (phones.moveToNext()) { 
// Get the phone number!? 

String contactName = phones.getString( 
    phones.getColumnIndex( 
    ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME)); 

String phoneNumber = phones.getString( 
    phones.getColumnIndex( 
    ContactsContract.CommonDataKinds.Phone.NUMBER)); 
Toast.makeText(this, phoneNumber, Toast.LENGTH_LONG).show();  

drawContact(contactName, phoneNumber); 

} 
phones.close(); 
    } 
    }cursor.close(); 
} 

private void drawContact(String name, String number){ 

    HashMap<String,String> item = new HashMap<String,String>(); 
    item.put("line1",name); 
    item.put("line2",number); 
    list.add(item); 
    numbers.notifyDataSetChanged(); 

} 

} 

It'seems沒有接觸有一個電話號碼(我已經添加在模擬器上2個觸點,我也試過在我的HTC Desire)。問題是, 如果(布爾.parseBoolean(hasPhone)) 返回始終爲false .. 我怎樣才能得到正確的電話號碼?

我試圖在if語句之前調用drawContact(String name,String number)而不查詢電話號碼,並且它工作(它繪製了兩次名稱)。但在LinearLayout上,它們不是按字母順序排列的...我如何按字母順序排序(類似於原始聯繫人應用程序)?

謝謝你的建議, 盧卡

回答

1

檢查documentation的parseBoolean()。 您可能想嘗試將結果解析爲int而不是布爾值。