MySQL查询展现连续的结果

发布时间:2019-04-02  栏目:sqlite  评论:0 Comments

#mysql中 对于查询结果只呈现n条一连行的标题#

在领扣上遭遇的贰个标题:求满意条件的连接3行结果的来得

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.首先先进行结果集的询问

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

二.给查询的结果集扩张二个自增列

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

叁.自增列和id的差值 相同即延续

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

4.将同样的差值 放在同样张表中,并取出三番五次数量超出三的

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

伍.将上步获得的表和主表 取得所须求的

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);

闻讯还是能够用存款和储蓄进度来成功,可是小编没尝试,稍后尝试

以上

  select CONCAT(‘My’, NULL, ‘QL’);

  再次回到来自于参数连结的字符串。假诺其余参数是NULL,
再次来到NULL。可以有超过一个的参数。二个数字参数被变换为等价的字符串格局。

  -> ‘14.3’

  [示例]

  】

  1、【CONCAT(str1,str2,…)

  从左初阶截取字符串.表达:left(被截取字段,截取长度)

  】

  2、【LEFT(str,length)

  select CONCAT(‘My’, ‘S’, ‘QL’);

  结合1、2 :concat ( left (数值1 / 数值2 *100,5),’%’) as 投诉率

   找了有个别素材,然后本人是用到了MySQL字符串处理中的五个函数concat()和left()

  -> NULL

  -> ‘MySQL’

http://www.bkjia.com/Mysql/434746.htmlwww.bkjia.comtruehttp://www.bkjia.com/Mysql/434746.htmlTechArticle找了一些资料,然后我是用到了MySQL字符串处理中的两个函数concat()和left()
1、【CONCAT(str一,str二,…)
再次回到来自于参数连结的字符串。要是别的参…

  select CONCAT(14.3);

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