MySQL查询显示连续的结果

2019-11-16 13:47栏目:计算机论坛
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#mysql中 对于查询结果只体现n条三番两次行的主题素材#

在领扣上碰到的叁个主题材料:求满足条件的总是3行结果的体现

X city built a new stadium, each day many people visit it and the stats are saved as these columns: id, date, people;
Please write a query to display the records which have 3 or more consecutive rows and the amount of people more than 100(inclusive).
For example, the table stadium:
+------+------------+-----------+
| id   | date       | people    |
+------+------------+-----------+
| 1    | 2017-01-01 | 10        |
| 2    | 2017-01-02 | 109       |
| 3    | 2017-01-03 | 150       |
| 4    | 2017-01-04 | 99        |
| 5    | 2017-01-05 | 145       |
| 6    | 2017-01-06 | 1455      |
| 7    | 2017-01-07 | 199       |
| 8    | 2017-01-08 | 188       |
+------+------------+-----------+

For the sample data above, the output is:
+------+------------+-----------+
| id   | date       | people    |
+------+------------+-----------+
| 5    | 2017-01-05 | 145       |
| 6    | 2017-01-06 | 1455      |
| 7    | 2017-01-07 | 199       |
| 8    | 2017-01-08 | 188       |
+------+------------+-----------+

1.率先先进行理并了结果集的查询

select id,date,people from stadium where people>=100;

2.给查询的结果集扩充一个自增列

SELECT @newid:=@newid+1 AS newid,test.* 
FROM(SELECT @newid:=0)r, test WHERE people>100

3.自增列和id的差值 相像即一而再一而再

SELECT @newid:=@newid+1 AS newid,test.* ,@cha:=id-@newid AS cha 
FROM(SELECT @newid:=0)r, test WHERE people>100

4.将生机勃勃律的差值 放在雷同张表中,并收取三番五次数量超过3的

select if(count(id)>=3,count_concat(id),null)e from(
SELECT @newid:=@newid+1 AS newid,test.* ,@cha:=id-@newid AS cha 
FROM(SELECT @newid:=0)r, test WHERE people>100)
as d group by cha

5.将上步获得的表和主表 获得所需求的

SELECT id,DATE,people FROM test,
(SELECT IF (COUNT(id)>3,GROUP_CONCAT(id),NULL)e 
FROM (SELECT @newid:=@newid+1 AS newid,test.* ,@cha:=id-@newid AS cha 
FROM(SELECT @newid:=0)r, test WHERE people>100)AS d   GROUP BY cha ) AS f 
WHERE f.e IS NOT NULL AND FIND_IN_SET(id,f.e);

据说还足以用存款和储蓄进程来成功,可是自个儿没尝试,稍后尝试

以上

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