1. 查重
   select*frompeoplewherepeopleIdin(selectpeopleIdfrompeoplegroupbypeopleIdhavingcount(peopleId)>1) select * from Users where UserId in (select UserId from Users group by UserId having count(UserId) > 1)

   不過在資料量過大的時候查詢的速度會非常慢

   
   selectcount(peopleId)frompeople selectcount(distinctpeopleId)frompeople
   我們可以用上面兩條語句的結果進行對比判斷是否存在重複資料
2. 分割槽間
   

   業務場景：統計消費總金額大於1000元的，800到1000元的，500到800元的，以及500元以下的人數。

   SELECT COUNT(CASE WHEN IFNULL(total_money,0) >=1000 THEN a.customer_id END) AS '>1000'
         ,COUNT(CASE WHEN IFNULL(total_money,0) >=800 AND IFNULL(total_money,0) <1000 THEN a.customer_id END) AS '800~1000'
         ,COUNT(CASE WHEN IFNULL(total_money,0) >=500 AND IFNULL(total_money,0) <800 THEN a.customer_id END) AS '500~800'
         ,COUNT(CASE WHEN IFNULL(total_money,0) <500  THEN a.customer_id END) AS '<500'
   FROM mc_userdb.`customer_login` a
   LEFT JOIN
   ( SELECT customer_id,SUM(order_money) AS total_money
     FROM mc_orderdb.`order_master` GROUP BY customer_id) b
   ON a.`customer_id`=b.`customer_id`

3. 現在要統計：<50、50-60、60-70、70-80、80-90、90-100、>=100分數區間的人數；
利用 INTERVAL 劃出7個區間；
再利用 elt 函式將7個區間分別返回一個列名，如下SQL：

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mysql> SELECT elt(INTERVAL(score, 0, 50, 60, 70, 80, 90, 100), '<50', '50-60', '60-70', '70-80', '80-90', '90-100', '>=100') as score_level, count(name) as counts
    -> FROM class
    -> GROUP BY elt(INTERVAL(score, 0, 50, 60, 70, 80, 90, 100), '<50', '50-60', '60-70', '70-80', '80-90', '90-100', '>=100');
+-------------+--------+
| score_level | counts |
+-------------+--------+
| 50-60       |      1 |
| 60-70       |      3 |
| 70-80       |      1 |
| 80-90       |      2 |
| 90-100      |      1 |
| <50         |      1 |
| >=100       |      1 |
+-------------+--------+
7 rows in set (0.00 sec)
 

4. update select用法

UPDATEuser_online_month_atu a