如何在一个SQLite查询中获取联系人ID,电子邮件,电话号码？联系人Android Optimization

我想通过一个电话号码获取所有联系人，我也想要每个联系人的所有电话号码和所有电子邮件。

当前代码：

// To get All Contacts having atleast one phone number.

Uri uri = ContactsContract.Contacts.CONTENT_URI;
String selection = ContactsContract.Contacts.HAS_PHONE_NUMBER + " > ?";
String[] selectionArgs = new String[] {"0"};
Cursor cu = applicationContext.getContentResolver().query(uri,null,selection,selectionArgs,null);

// For getting All Phone Numbers and Emails further queries : 
while(cu.movetoNext()){
String id = cu.getString(cu.getColumnIndex(ContactsContract.Contacts._ID));


 // To get Phone Numbers of Contact
    Cursor pCur = context.getContentResolver().query(
    ContactsContract.CommonDataKinds.Phone.CONTENT_URI,ContactsContract.CommonDataKinds.Phone.CONTACT_ID + "=?",new String[]{id},null);

// To get Email ids of Contact
Cursor emailCur = context.getContentResolver().query(
ContactsContract.CommonDataKinds.Email.CONTENT_URI,ContactsContract.CommonDataKinds.Email.CONTACT_ID + " = ?",null); 

// Iterate through these cursors to get Phone numbers and Emails
}

如果我的设备中有超过1000个联系人，则需要花费太多时间。如何在单个查询中获取所有数据，而不是为每个联系人增加两个查询？

还是有其他的优化方法？

先谢谢你。

ICS：当您从Data.CONTENT_URI查询时，您已将所有相关联的联系人的行已加入 – 即这将工作：

ContentResolver resolver = getContentResolver();
Cursor c = resolver.query(
        Data.CONTENT_URI,Data.HAS_PHONE_NUMBER + "!=0 AND (" + Data.MIMETYPE + "=? OR " + Data.MIMETYPE + "=?)",new String[]{Email.CONTENT_ITEM_TYPE,Phone.CONTENT_ITEM_TYPE},Data.CONTACT_ID);

while (c.movetoNext()) {
    long id = c.getLong(c.getColumnIndex(Data.CONTACT_ID));
    String name = c.getString(c.getColumnIndex(Data.disPLAY_NAME));
    String data1 = c.getString(c.getColumnIndex(Data.DATA1));

    System.out.println(id + ",name=" + name + ",data1=" + data1);
}

如果您要定位2.3，则需要考虑到在查询数据时使用的连接不能使用HAS_PHONE_NUMBER。

乐趣。

例如，这可以通过跳过您的联系人必须具有电话号码的要求来解决，而不是解决“至少有一个电话号码或电子邮件地址的任何联系人”：

Cursor c = resolver.query(
        Data.CONTENT_URI,Data.MIMETYPE + "=? OR " + Data.MIMETYPE + "=?",Data.CONTACT_ID);

如果这不是一个选项，你总是可以去一个可怕的黑客子选择：

Cursor c = resolver.query(
        Data.CONTENT_URI,"(" + Data.MIMETYPE + "=? OR " + Data.MIMETYPE + "=?) AND " + 
        Data.CONTACT_ID + " IN (SELECT " + Contacts._ID + " FROM contacts WHERE " + Contacts.HAS_PHONE_NUMBER + "!=0)",Data.CONTACT_ID);

或使用两个游标来解决它：

Cursor contacts = resolver.query(Contacts.CONTENT_URI,Contacts.HAS_PHONE_NUMBER + " != 0",Contacts._ID + " ASC");
Cursor data = resolver.query(Data.CONTENT_URI,Data.CONTACT_ID + " ASC");

int idindex = contacts.getColumnIndexOrThrow(Contacts._ID);
int nameIndex = contacts.getColumnIndexOrThrow(Contacts.disPLAY_NAME);
int cidindex = data.getColumnIndexOrThrow(Data.CONTACT_ID);
int data1Index = data.getColumnIndexOrThrow(Data.DATA1);
boolean hasData = data.movetoNext();

while (contacts.movetoNext()) {
    long id = contacts.getLong(idindex);
    System.out.println("Contact(" + id + "): " + contacts.getString(nameIndex));
    if (hasData) {
        long cid = data.getLong(cidindex);
        while (cid <= id && hasData) {
            if (cid == id) {
                System.out.println("\t(" + cid + "/" + id + ").data1:" + 
                        data.getString(data1Index));
            }
            hasData = data.movetoNext();
            if (hasData) {
                cid = data.getLong(cidindex);
            }
        }
    }
}

原文地址：https://www.jb51.cc/sqlite/197968.html