我正在尝试从手机中检索只有数字的联系人并将它们放入arrayList中,在懒惰的适配器中查看它们,点击名称我只想显示数字。我设法获得了联系人和号码列表,但问题是当我有多个号码的联系人时,它只是加入到列表中。
类似于例如
大卫+1 508 656 9043
大卫+1 403 604 7053
大卫+1 212 608 7053
相反,我想在列表中仅显示David,当我点击它时应显示所有三个数字。
我试过了:
void getContactNumbers()
{
ContentResolver cr = getContentResolver();
String sortOrder = ContactsContract.Contacts.DISPLAY_NAME+ " COLLATE LOCALIZED ASC";
Cursor cur = cr.query(ContactsContract.Contacts.CONTENT_URI, null, null, null, sortOrder);
if (cur.getCount() > 0)
{
while (cur.moveToNext())
{
String id = cur.getString(cur.getColumnIndex(ContactsContract.Contacts._ID));
String name = cur.getString(cur.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
Log.e("contact", "...Contact Name ...." + name);
if (Integer.parseInt(cur.getString(cur.getColumnIndex(ContactsContract.Contacts.HAS_PHONE_NUMBER))) > 0)
{
Cursor pCur = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI,null,ContactsContract.CommonDataKinds.Phone.CONTACT_ID +" = ?",
new String[]{id}, null);
while (pCur.moveToNext())
{
String phoneNo = pCur.getString(pCur.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
Log.e("contact", "...Contact Name ...." + name + "...contact Number..." + phoneNo);
}
pCur.close();
}
}
}
}
如何解决这一部分?
谢谢!
答案 0 :(得分:3)
谢谢Harshid。选择更改而不是IN_VISIBLE_GROUP +“='1'”; - 我添加了HAS_PHONE_NUMBER +“='1'”;
所有联系人都出现了..希望以下代码可以帮助其他人!
谢谢!
final Uri uri = ContactsContract.Contacts.CONTENT_URI;
final String[] projection = new String[] {
ContactsContract.Contacts._ID,
ContactsContract.Contacts.DISPLAY_NAME,
ContactsContract.Contacts.PHOTO_ID
};
String selection = ContactsContract.Contacts.HAS_PHONE_NUMBER + " = '1'";
final String sortOrder = ContactsContract.Contacts.DISPLAY_NAME + " COLLATE LOCALIZED ASC";
Cursor cur = getContentResolver().query(uri, projection, selection, null, sortOrder);
if (cur.getCount() > 0)
{
while (cur.moveToNext())
{
String Sid = cur.getString(cur.getColumnIndex(ContactsContract.Contacts._ID));
String name = cur.getString(cur.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
Log.e("contact", "...Contact Name ...." + name);
// get the phone number
Cursor pCur = getApplicationContext().getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null,ContactsContract.CommonDataKinds.Phone.CONTACT_ID
+ " = ?", new String[] { id }, null);
while (pCur.moveToNext()) {
number = pCur.getString(pCur .getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
}
pCur.close();
}
}
cur.close();
答案 1 :(得分:-1)
您必须这样查询并get contact
与multiple number
。
final Uri uri = ContactsContract.Contacts.CONTENT_URI;
final String[] projection = new String[] {
ContactsContract.Contacts._ID,
ContactsContract.Contacts.DISPLAY_NAME,
ContactsContract.Contacts.PHOTO_URI
};
String selection = ContactsContract.Contacts.IN_VISIBLE_GROUP + " = '1'";
String[] selectionArgs = null;
final String sortOrder = ContactsContract.Contacts.DISPLAY_NAME + " COLLATE LOCALIZED ASC";
Cursor cur = getContentResolver().query(uri, projection, selection, selectionArgs, sortOrder);
if (cur.getCount() > 0)
{
while (cur.moveToNext())
{
String id = cur.getString(cur.getColumnIndex(ContactsContract.Contacts._ID));
String name = cur.getString(cur.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME));
Log.e("contact", "...Contact Name ...." + name);
// get the phone number
Cursor pCur = getApplicationContext().getContentResolver().query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, null,ContactsContract.CommonDataKinds.Phone.CONTACT_ID
+ " = ?", new String[] { id }, null);
while (pCur.moveToNext()) {
number = pCur.getString(pCur .getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER));
}
pCur.close();
}
}
cur.close();
如果getting error
请尝试此代码,然后将评论另外享受..