仅从Android联系人中获取电子邮件联系人 [英] Fetch only Email contacts from Android Contacts
问题描述
我正在尝试仅获取联系人中可用的电子邮件联系人。现在,我有了一个可以显示所有联系人的解决方案,如果选定的联系人没有电子邮件地址,它将敬酒,说明找不到电子邮件地址。相反,我想显示仅具有电子邮件地址的联系人。
I am trying to fetch only email contacts that are available in my contacts. Right now I have got a solution that shows all the contacts and if selected contacts doesn't have an email address it would toast stating no email address found. Instead I would like to show contacts that has only email address.
这是我尝试的查询:
Cursor cursor = null;
String emailid = "";
List<String> allids = new ArrayList<String>();
int emailIds = 0;
try
{
Uri result = data.getData();
String id = result.getLastPathSegment();
Log.e("Email","TRY"+emailid);
cursor = getContentResolver().query(Email.CONTENT_URI, null, Email.CONTACT_ID + "=?", new String[] { id }, null);
/*cursor = getContentResolver().query(ContactsContract.CommonDataKinds.Email.CONTENT_URI, null, ContactsContract.CommonDataKinds.Email.CONTACT_ID + " = ?",
new String[]{id}, null);*/
emailIds = cursor.getColumnIndex(Email.DATA);
if (cursor.moveToFirst())
{
while (cursor.isAfterLast() == false)
{
emailid = cursor.getString(emailIdx);
allids.add(emailid);
cursor.moveToNext();
}
}
else
{
//no results actions
}
}
有人可以让我知道如何使电子邮件查询部分正常工作吗?
Can somebody let me know how to get email query part working?
谢谢!
推荐答案
2通过电子邮件获取联系人的算法
2 Algorithm that fetches contact with email
public ArrayList<String> getNameEmailDetails(){
ArrayList<String> names = new ArrayList<String>();
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._ID));
Cursor cur1 = cr.query(
ContactsContract.CommonDataKinds.Email.CONTENT_URI, null,
ContactsContract.CommonDataKinds.Email.CONTACT_ID + " = ?",
new String[]{id}, null);
while (cur1.moveToNext()) {
//to get the contact names
String name=cur1.getString(cur1.getColumnIndex(ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME));
Log.e("Name :", name);
String email = cur1.getString(cur1.getColumnIndex(ContactsContract.CommonDataKinds.Email.DATA));
Log.e("Email", email);
if(email!=null){
names.add(name);
}
}
cur1.close();
}
}
return names;
}
上述方法返回具有电子邮件ID的名称数组列表。
但是很慢,这是另一个算法,速度更快:
the above method return an arraylist of names which has email id. But is slow, here is an another algorithm, much faster:
public ArrayList<String> getNameEmailDetails() {
ArrayList<String> emlRecs = new ArrayList<String>();
HashSet<String> emlRecsHS = new HashSet<String>();
Context context = getActivity();
ContentResolver cr = context.getContentResolver();
String[] PROJECTION = new String[] { ContactsContract.RawContacts._ID,
ContactsContract.Contacts.DISPLAY_NAME,
ContactsContract.Contacts.PHOTO_ID,
ContactsContract.CommonDataKinds.Email.DATA,
ContactsContract.CommonDataKinds.Photo.CONTACT_ID };
String order = "CASE WHEN "
+ ContactsContract.Contacts.DISPLAY_NAME
+ " NOT LIKE '%@%' THEN 1 ELSE 2 END, "
+ ContactsContract.Contacts.DISPLAY_NAME
+ ", "
+ ContactsContract.CommonDataKinds.Email.DATA
+ " COLLATE NOCASE";
String filter = ContactsContract.CommonDataKinds.Email.DATA + " NOT LIKE ''";
Cursor cur = cr.query(ContactsContract.CommonDataKinds.Email.CONTENT_URI, PROJECTION, filter, null, order);
if (cur.moveToFirst()) {
do {
// names comes in hand sometimes
String name = cur.getString(1);
String emlAddr = cur.getString(3);
// keep unique only
if (emlRecsHS.add(emlAddr.toLowerCase())) {
emlRecs.add(emlAddr);
}
} while (cur.moveToNext());
}
cur.close();
return emlRecs;
}
第一个代码花了大约4秒钟才能在我的测试设备上获取联系人,第二个代码花费了大约0.04秒。
first code took about 4 seconds to get contacts on my test device, and this second code took about 0.04 sec.
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