Question

Play商店中有许多应用程序，这些应用程序在每个用户的末尾都显示联系人（在征得用户许可后）和公司徽标，这表示用户是否在该应用程序上。此流程有助于邀请一组用户。当我实现上述所有功能时，事情变得非常复杂，假设底部导航器有一个专门的操作 Contacts ，当我从已知方法（Content Resolver查询）中获取联系人时，首先它太慢了，因此我将联系人添加到sqlite数据库，但仍然很慢。 如何有效地提取联系人？。像 OYO ROOMS APP 一样，它具有专门的Bottom Nav Action Invite＆earn ，该功能正是我想要的。如果有人知道如何有效地做到这一点，请告诉我。

private void fetchContactsFromSystem() {
    ContentResolver cr = getContext().getContentResolver();
    Cursor cursor = cr.query(ContactsContract.CommonDataKinds.Phone.CONTENT_URI, PROJECTION, null, null, ContactsContract.CommonDataKinds.Phone.DISPLAY_NAME + " ASC");
    if (cursor != null) {
        try {
            final int nameIndex = cursor.getColumnIndex(ContactsContract.Contacts.DISPLAY_NAME);
            final int numberIndex = cursor.getColumnIndex(ContactsContract.CommonDataKinds.Phone.NUMBER);
            String name, number;
            while (cursor.moveToNext()) {
                name = cursor.getString(nameIndex);
                number = cursor.getString(numberIndex);
                number = number.replaceAll(" ", "");
                if (!phoneNumberSet.contains(number.trim())) {
                    sqliteDatabaseHelper.insertData(name.trim(), number.trim());
                    Contact contact = new Contact();
                    contact.setName(name);
                    contact.setPhone(number);
                    contact.setInvited(false);
                    contactList.add(contact);
                    phoneNumberSet.add(number);
                }
            }
        } finally {
            cursor.close();
        }
    }
}

投影字符串数组

private static final String[] PROJECTION = new String[]{
        ContactsContract.CommonDataKinds.Phone.CONTACT_ID,
        ContactsContract.Contacts.DISPLAY_NAME,
        ContactsContract.CommonDataKinds.Phone.NUMBER
};

从sqlite获取数据

    private void setUpFromSqliteDatabase(Cursor cr) {
    String name, number;
    boolean isInvite;
    while (cr.moveToNext()) {
        name = cr.getString(0);
        number = cr.getString(1);
        isInvite = cr.getInt(2) == 1;
        Contact contact = new Contact();
        contact.setName(name);
        contact.setPhone(number);
        contact.setInvited(isInvite);
        contactList.add(contact);
    }
    cr.close();
}

SQLite助手类

public class SqliteDatabaseHelper extends SQLiteOpenHelper {
private static final String DATABASE_NAME = "contacts.db";
private static final String TABLE_NAME = "contacts";
public static final String NAME = "NAME";
public static final String PHONE_NUMBER = "PHONE_NUMBER";
public static final String IS_INVITED = "IS_INVITED";

public SqliteDatabaseHelper(Context context) {
    super(context, DATABASE_NAME, null, 1);
}

@Override
public void onCreate(SQLiteDatabase db) {
    db.execSQL("create table " + TABLE_NAME +" (NAME TEXT, PHONE_NUMBER TEXT PRIMARY KEY, IS_INVITED INTEGER DEFAULT 0)");
}

@Override
public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    db.execSQL("DROP TABLE IF EXISTS "+TABLE_NAME);
    onCreate(db);
}

public boolean insertData(String name,String phone) {
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues contentValues = new ContentValues();
    contentValues.put(NAME,name);
    contentValues.put(PHONE_NUMBER,phone);
    long result = db.insert(TABLE_NAME,null ,contentValues);
    db.close();
    return result != -1;
}

public void updateInvitedData(String phoneNumber) {
    SQLiteDatabase db = this.getWritableDatabase();
    ContentValues cv = new ContentValues();
    cv.put(IS_INVITED, 1);
    db.update(TABLE_NAME, cv, "PHONE_NUMBER = ?",new String[] {phoneNumber});
    db.close();
}

public Cursor getAllData() {
    SQLiteDatabase db = this.getWritableDatabase();
    return db.rawQuery("select * from "+TABLE_NAME,null);
}

}

Answer 1

有几个小区域可以提高效率，主要的购买要分批处理插件。这是一个示例，说明如何修改现有代码以实现此目的。

private void fetchContactsFromSystem() {
    final ContentResolver cr = getContext().getContentResolver();
    final String sort = String.format("%s ASC", Phone.DISPLAY_NAME);
    try(Cursor cursor = cr.query(Phone.CONTENT_URI, PROJECTION, null, null, sort)) {
        if(cursor == null || !cursor.moveToFirst()) return;

        final List<Contact> insert = new LinkedList<>();
        final int nameIndex = cursor.getColumnIndex(Contacts.DISPLAY_NAME);
        final int numberIndex = cursor.getColumnIndex(Phone.NUMBER);

        do {
            final String name = cursor.getString(nameIndex);
            final String number = cursor.getString(numberIndex).replaceAll("\\s+","");

            //"true" if number doesn't already exist in the Set
            if(phoneNumberSet.add(number)) {
                //Can make this immutable and pass info in constructor
                final Contact contact = new Contact(name, number, false);
                contactList.add(contact);
                insert.add(contact);  
            }       
        } while(cursor.moveToNext());
        //Conduct bulk insert of "insert" list here
        bulkInsert(insert);
    }

}

private void bulkInsert(List<Contact> contacts) {
    final SQLiteDatabase db = sqliteDatabaseHelper.getWritableDatabase();
    db.beginTransaction();
    try {
        for (Contact contact : contacts) {
            final ContentValues cv = new ContentValues(2);
            cv.put(NAME_ID, contact.name());
            cv.put(NUMBER_ID, contact.number());
            db.insert(TABLE_CONTACTS, null, values);
        }
        db.setTransactionSuccessful();
    } finally {
        db.endTransaction();
    }
}

Answer 2

如果预期的结果是一组电话，则不需要按DISPLAY_NAME进行排序，那么SQLite中的排序通常是一项昂贵的功能。

此外，我不会将结果存储在本地SQLite中，当尝试保持本地副本和更新的Contacts DB同步时（例如，当用户添加或删除联系人时），它将导致您出现问题。

如何查看应用程序中存在的联系人？

2 个答案: