4

我正在使用 open Id login 审核我的应用程序中的用户详细信息。如果用户第一次登录 OPEN ID，我们将其视为注册。我正在使用此详细信息生成审核登录报告。样本表数据。

+---------+----------+-----------+---------------+
| USER_ID | PROVIDER | OPERATION | TIMESTAMP     |
+---------+----------+-----------+---------------+
|     120 | Google   | SIGN_UP   | 1347296347000 |
|     120 | Google   | SIGN_IN   | 1347296347000 |
|     121 | Yahoo    | SIGN_IN   | 1347296347000 |
|     122 | Yahoo    | SIGN_IN   | 1347296347000 |
|     120 | Google   | SIGN_UP   | 1347296347000 |
|     120 | FaceBook | SIGN_IN   | 1347296347000 |
+---------+----------+-----------+---------------+

在此表中，我想根据提供者排除已经SIGN_UP ed " SIGN_IN " ed 的用户计数。

显示创建表

CREATE TABLE `signin_details` (
  `USER_ID` int(11) DEFAULT NULL,
  `PROVIDER` char(40) DEFAULT NULL,
  `OPERATION` char(40) DEFAULT NULL,
  `TIMESTAMP` bigint(20) DEFAULT NULL
) ENGINE=InnoDB

我正在使用这个查询。 

select 
  count(distinct(USER_ID)) as signin_count, 
  PROVIDER from signin_details s1 
where 
  s1.USER_ID NOT IN 
  (
    select 
      USER_ID 
    from signin_details 
    where 
      signin_details.PROVIDER=s1.PROVIDER 
      and signin_details.OPERATION='SIGN_UP' 
      and signin_details.TIMESTAMP/1000 BETWEEN UNIX_TIMESTAMP(CURRENT_DATE()-INTERVAL 1 DAY) * 1000 AND UNIX_TIMESTAMP(CURRENT_DATE()) * 1000
  )  
  AND OPERATION='SIGN_IN' group by PROVIDER;

解释输出：

+----+--------------------+----------------+------+---------------+------+---------+------+------+-----------------------------+
| id | select_type        | table          | type | possible_keys | key  | key_len | ref  | rows | Extra                       |
+----+--------------------+----------------+------+---------------+------+---------+------+------+-----------------------------+
|  1 | PRIMARY            | s1             | ALL  | NULL          | NULL | NULL    | NULL |    6 | Using where; Using filesort |
|  2 | DEPENDENT SUBQUERY | signin_details | ALL  | NULL          | NULL | NULL    | NULL |    6 | Using where                 |
+----+--------------------+----------------+------+---------------+------+---------+------+------+-----------------------------+

查询输出：

+--------------+----------+
| signin_count | PROVIDER |
+--------------+----------+
|            1 | FaceBook |
|            2 | Yahoo    |
+--------------+----------+

执行 200k 行需要 40 多分钟。

我的假设是它将检查每一行的依赖子查询输出的总数。

我对此查询的假设。

 A -> Dependant Outputs (B,C,D) .
 A check with B
 A check with C
 A check with D

如果相关查询输出较大，则执行时间会很长。如何改进这个查询？

4

2 回答 2

4

如果您使用MySQL，您必须知道子查询的执行速度非常慢。

IN是缓慢的...

EXISTS然后通常更快IN

JOIN主要是做这种事情的最快方法。

SELECT DISTINCT
  s1.PROVIDER,
  COUNT(DISTINCT s1.USER_ID)

FROM 
  signin_details s1
  LEFT JOIN 
  (
    SELECT DISTINCT
      USER_ID, PROVIDER
    FROM 
      signin_details 
    WHERE
      signin_details.OPERATION='SIGN_UP' 
      AND 
        signin_details.TIMESTAMP 
          BETWEEN 
            UNIX_TIMESTAMP(CURRENT_DATE()-INTERVAL 1 DAY) * 1000 
            AND UNIX_TIMESTAMP(CURRENT_DATE()) * 1000
  ) AS t USING  (USER_ID, PROVIDER)

WHERE
  t.USER_ID IS NULL
  AND OPERATION='SIGN_IN'
GROUP BY s1.PROVIDER

http://sqlfiddle.com/#!2/122ac/12

注意：如果您想知道 sqlfiddle 结果，请考虑这里是UNIX_TIMESTAMP查询中的一个。

结果：

| PROVIDER | COUNT(DISTINCT S1.USER_ID) |
-----------------------------------------
| FaceBook |                          1 |
|    Yahoo |                          2 |

MySQL 和INTERSECT故事。你得到所有你不想计算USER_ID的组合。PROVIDER然后LEFT JOIN他们到你的数据。现在，您要计算的所有行都没有来自LEFT JOIN. 你得到他们t.USER_ID IS NULL。

输入：

| rn° | USER_ID | PROVIDER | OPERATION |     TIMESTAMP |
-------------------------------------------------------
| 1   |     120 |   Google |   SIGN_UP | 1347296347000 | -
| 2   |     120 |   Google |   SIGN_IN | 1347296347000 | - (see rn° 1)
| 3   |     121 |    Yahoo |   SIGN_IN | 1347296347000 | Y
| 4   |     122 |    Yahoo |   SIGN_IN | 1347296347000 | Y
| 5   |     120 |   Google |   SIGN_UP | 1347296347000 | -
| 6   |     120 | FaceBook |   SIGN_IN | 1347296347000 | F
| 7   |     119 | FaceBook |   SIGN_IN | 1347296347000 | - (see rn° 8)
| 8   |     119 | FaceBook |   SIGN_UP | 1347296347000 | -
于 2012-09-10T18:31:42.817 回答
0

在 HAVING 子句中使用“NOT IN”。它会比“不在哪里”更快

于 2019-11-03T06:32:58.410 回答