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MySQL查询显示连续的结果香港六马会开奖结果
分类:编程教程

#mysql中 对于查询结果只展现n条接二连三行的标题#

在领扣上遇见的三个难点:求满意条件的连年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.第黄金年代先实行理并了结果集的询问

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

2.给查询的结果集增添二个自增列

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 雷同即三回九转

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

4.将长久以来的差值 放在同等张表中,并抽取延续数量超出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.将上步得到的表和主表 得到所供给的

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);

闻讯还能用存款和储蓄过程来形成,可是小编没尝试,稍后尝试

以上

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