Question

这就是我已经拥有的。

var bookRef = firebaseRef.database().ref('path/to/books/book1');

bookRef.on('value', function(snapshot) {
    var bookData = snapshot.val();
    var newData = {};
    var newTitle = 'NewTitle';

    bookData.bookInfo.bookTitle = newTitle;
    newData[newTitle] = bookData;
    firebaseRef.database().ref('path/to/books/' + newTitle).set(newData);

});

如何按SELECT karma , profanity , username FROM users ORDER BY (karma - profanity) DESC LIMIT 10订购和ORDER BY (karma - profanity) DESC LIMIT 10

ORDER BY profanity DESC LIMIT 10

所需结果将如下所示：

CREATE TABLE Test
    (`id` int, `username` varchar(55), `karma` int,`profanity` int)
;

INSERT INTO Test
    (`id`, `username`, `karma`, `profanity`)
VALUES
    (1, 'User1', '10', '1'),
    (2, 'User2', '8', '2'),
    (3, 'User3', '1', '2'),
    (4, 'User4', '11', '1'),
    (5, 'User5', '5', '0'),
    (6, 'User6', '6', '3'),
    (7, 'User7', '1', '1'),
    (8, 'User8', '2', '3'),
    (9, 'User9', '2', '1'),
    (10, 'User10', '1', '7'),
    (11, 'User11', '7', '7'),
    (12, 'User12', '1', '1'),
    (13, 'User13', '10', '0'),
    (14, 'User14', '1', '3'),
    (15, 'User15', '7', '0')
;

都有输出karma, profanity, username | profanity username 10 0 User13 7 User11 11 1 User4 7 User10 10 1 User1 3 User8 7 0 User15 3 User6 8 2 User2 3 User14 5 0 User5 2 User3 6 3 User6 2 User2 2 1 User9 1 User4 1 1 User7 1 User7 1 1 User12 1 User9

两种排序合并为一个具有不同标注顺序的订单

http://sqlfiddle.com/#!9/7ca828/2

Answer 1

我认为这是转向显示问题 - 您通常在应用程序级代码中解决的问题，但无论如何......

SELECT a.karma a_karma
     , a.profanity a_profanity
     , a.username a_username
     , b.profanity b_profanity
     , b.username b_username
  FROM
     ( SELECT *,@kp:=@kp+1 kp FROM test, (SELECT @kp:=0) vars ORDER BY karma-profanity DESC LIMIT 10 ) a
  JOIN
     ( SELECT *,@p:=@p+1 p FROM test, (SELECT @p:=0) vars ORDER BY profanity DESC LIMIT 10 ) b
    ON b.p = a.kp;
+---------+-------------+------------+-------------+------------+
| a_karma | a_profanity | a_username | b_profanity | b_username |
+---------+-------------+------------+-------------+------------+
|      10 |           0 | User13     |           7 | User11     |
|      11 |           1 | User4      |           7 | User10     |
|      10 |           1 | User1      |           3 | User8      |
|       7 |           0 | User15     |           3 | User6      |
|       8 |           2 | User2      |           3 | User14     |
|       5 |           0 | User5      |           2 | User3      |
|       6 |           3 | User6      |           2 | User2      |
|       2 |           1 | User9      |           1 | User4      |
|       1 |           1 | User7      |           1 | User7      |
|       1 |           1 | User12     |           1 | User9      |
+---------+-------------+------------+-------------+------------+

Mysql查询顺序由多个项目（声明）

1 个答案: