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我发现这个漂亮的联系人列表布局：https ://github.com/thehung111/ContactListView

但是，联系人是硬编码的。所以我需要获取电话联系人并填写联系人列表。

这是我尝试过的：

public class ExampleDataSource  {

public static List<ContactItemInterface> getSampleContactList(){
     List<ContactItemInterface>  list = new  ArrayList<ContactItemInterface> ();


     Uri uri = ContactsContract.CommonDataKinds.Phone.CONTENT_URI;
    String[] projection    = new String[] {ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME,
                         ContactsContract.CommonDataKinds.Phone.NUMBER};
     Cursor people = getContentResolver().query(uri, projection, null, null, null);

     int indexName = people.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME);
     int indexNumber = people.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER);

     people.moveToFirst();
     do {
         String name   = people.getString(indexName);
         String number = people.getString(indexNumber);
         list.add(new ExampleContactItem(name , number ) );

     } while (people.moveToNext());

     /* Example inputs for contact list

     list.add(new ExampleContactItem("Lizbeth" , "Lizbeth Crockett" ) );
     list.add(new ExampleContactItem("Lizbeth" , "Lizbeth Crockett" ) );
     list.add(new ExampleContactItem("Zachery" , "Zachery Loranger" ) );
     list.add(new ExampleContactItem("Vada" , "Vada Winegar" ) );
     list.add(new ExampleContactItem("Essie" , "Essie Pass" ) );

    */   
     return list;
}

}

我在 getContentResolver() 上遇到错误，并试图将类扩展到应用程序等。到目前为止还没有运气。

所以主要问题是如何在Android上获取一个包含姓名和电话号码的列表作为字符串列表。

4

2 回答 2

0

您是否在清单中声明了以下权限？如果没有，请声明。

  <uses-permission android:name="android.permission.READ_CONTACTS" />
于 2013-10-15T13:05:21.223 回答
0

它将在新行中显示所有联系人姓名及其电话号码：

String names="";

Cursor people = getContentResolver().query(ContactsContract.Contacts.CONTENT_URI,
                                        null,null,null,
                                          null);

                                    while (people.moveToNext()) {


                                        int i=people.getColumnIndex(PhoneLookup.DISPLAY_NAME);
                                        int i2=people.getColumnIndex(PhoneLookup._ID);
                                        String name=people.getString(i);
                                        String id1=people.getString(i2);
                                        names+=name+":";
                                        Uri uri2 = ContactsContract.
                                                CommonDataKinds.Phone.CONTENT_URI;
                                                String[] projectio = new String[] {
                                                ContactsContract.CommonDataKinds.
                                                Phone.NUMBER }; 
                                                String selectio= ContactsContract.
                                                CommonDataKinds.Phone.CONTACT_ID +
                                                "=?";
                                                String[] selectionArg = new String[]
                                                {id1 };
                                                Cursor peopl=getContentResolver().query(uri2,null,selectio,selectionArg,null);
                                                while(peopl.moveToNext())   
                                                {
                                                    int i3=peopl.getColumnIndex(CommonDataKinds.Phone.NUMBER);

                                                    String phonenum=peopl.getString(i3);
                                                    names+=phonenum+"\n";


                                                }

                                             }
Log.d("names":names);
于 2015-08-28T07:04:20.360 回答