我从本地地址簿获取所有联系人,以及他们的姓名+电话号码,但问题是我与之交换邮件的返回电子邮件地址以及Skype ID,我不需要的东西: )
public void fetchContacts() {
String phoneNumber = null;
String email = null;
Uri CONTENT_URI = ContactsContract.Contacts.CONTENT_URI;
String _ID = ContactsContract.Contacts._ID;
String DISPLAY_NAME = ContactsContract.Contacts.DISPLAY_NAME;
String HAS_PHONE_NUMBER = ContactsContract.Contacts.HAS_PHONE_NUMBER;
Uri PhoneCONTENT_URI = ContactsContract.CommonDataKinds.Phone.CONTENT_URI;
String Phone_CONTACT_ID = ContactsContract.CommonDataKinds.Phone.CONTACT_ID;
String NUMBER = ContactsContract.CommonDataKinds.Phone.NUMBER;
Uri EmailCONTENT_URI = ContactsContract.CommonDataKinds.Email.CONTENT_URI;
String EmailCONTACT_ID = ContactsContract.CommonDataKinds.Email.CONTACT_ID;
String DATA = ContactsContract.CommonDataKinds.Email.DATA;
StringBuffer output = new StringBuffer();
ContentResolver contentResolver = getContentResolver();
Cursor cursor = contentResolver.query(CONTENT_URI, null, null, null, null);
// Loop for every contact in the phone
if (cursor.getCount() > 0) {
while (cursor.moveToNext()) {
String contact_id = cursor.getString(cursor.getColumnIndex(_ID));
String name = cursor.getString(cursor.getColumnIndex(DISPLAY_NAME));
int hasPhoneNumber = Integer.parseInt(cursor.getString(cursor.getColumnIndex(HAS_PHONE_NUMBER)));
if (hasPhoneNumber > 0) {
output.append("\n Name:" + name);
// Query and loop for every phone number of the contact
Cursor phoneCursor = contentResolver.query(PhoneCONTENT_URI, null, Phone_CONTACT_ID + " = ?", new String[]{contact_id}, null);
while (phoneCursor.moveToNext()) {
phoneNumber = phoneCursor.getString(phoneCursor.getColumnIndex(NUMBER));
output.append("\n Phone number:" + phoneNumber);
}
phoneCursor.close();
// Query and loop for every email of the contact
}
output.append("\n");
}
outputText.setText(output);
}
<TextView
android:id="@+id/textView1"
android:layout_width="wrap_content"
android:layout_height="wrap_content" />
答案 0 :(得分:1)
您的问题与Android无关,而是Java问题。
在最大的while
循环中,确保continue
IF中的hasPhoneNumber
关键字,或者只是将output.append("\n")
移到IF中。
干杯。 : - )