我正在尝试从电话号码中获取正确的联系人姓名,以将其显示为ListView
。
但我每次只获得所有电话号码的相同联系人姓名......
这是我的代码:
private String getContactName(String string){
String name=null;
ContentResolver cr=getContentResolver();
Cursor cur=cr.query(ContactsContract.Contacts.CONTENT_URI,null,null,null,null);
if(cur.getCount()>0){
while(cur.moveToNext()){
String id=cur.getString(cur.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER));
name=cur.getString(cur.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
}
}
return name;
}
public void onClick(View v)
{
ContentResolver contentResolver=getContentResolver();
Cursor cursor=contentResolver.query(Uri.parse("content://sms/inbox"),null,null,null,null);
int indexBody=cursor.getColumnIndex(SmsReceiver.BODY);
int indexAddr=cursor.getColumnIndex(SmsReceiver.ADDRESS);
// int indexAddr = Integer.parseInt(getContactName(Integer.toString(cursor.getColumnIndex( SmsReceiver.ADDRESS ))));
//getContactName(Integer.toString(indexAddr));
////
if(indexBody<0||!cursor.moveToFirst())return;
smsList.clear();
do
{
//String str = "Sender: " + cursor.getString( indexAddr ) + "\n" + cursor.getString( indexBody );
String str="Sender: "+getContactName(cursor.getString(indexAddr))+"\n"+cursor.getString(indexBody);
smsList.add(str);
}
while(cursor.moveToNext());
ListView smsListView=(ListView)findViewById(R.id.SMSList);
registerForContextMenu(smsListView);
smsListView.setAdapter(new ArrayAdapter<String>(this,android.R.layout.simple_list_item_1,smsList));
smsListView.setOnItemClickListener(this);
}
我找不到我的错误在哪里...并且只为所有发件人号码获取相同的名称。
谢谢
答案 0 :(得分:1)
以下是您的功能的正确代码。
private String getContactName(String phoneNumber)
{
String name = null;
ContentResolver cr = getContentResolver();
Cursor cur = cr.query(Uri.withAppendedPath(PhoneLookup.CONTENT_FILTER_URI,
Uri.encode(phoneNumber)), new String[]{ PhoneLookup.DISPLAY_NAME }, null, null, null);
if(cur != null){
try{
if(cur.getCount() > 0){
if(cur.moveToFirst()){
name = cur.getString(cur.getColumnIndex(PhoneLookup.DISPLAY_NAME));
}
}
}
finally{
cur.close();
}
}
return name;
}