基础篇
//查询时间，友好提示$sql = "select date_format(create_time, ‘%Y-%m-%d‘) as day from table_name";//int 时间戳类型$sql = "select from_unixtime(create_time, ‘%Y-%m-%d‘) as day from table_name";//一个sql返回多个总数$sql = "select count(*) all, " ; $sql .= " count(case when status = 1 then status end) status_1_num, "; $sql .= " count(case when status = 2 then status end) status_2_num "; $sql .= " from table_name";//Update Join / Delete Join$sql = "update table_name_1 "; $sql .= " inner join table_name_2 on table_name_1.id = table_name_2.uid "; $sql .= " inner join table_name_3 on table_name_3.id = table_name_1.tid "; $sql .= " set *** = *** "; $sql .= " where *** ";//delete join 同上。//替换某字段的内容的语句$sql = "update table_name set content = REPLACE(content, ‘aaa‘, ‘bbb‘) "; $sql .= " where (content like ‘%aaa%‘)";//获取表中某字段包含某字符串的数据$sql = "SELECT * FROM `表名` WHERE LOCATE(‘关键字‘, 字段名) ";//获取字段中的前4位$sql = "SELECT SUBSTRING(字段名,1,4) FROM 表名 ";//查找表中多余的重复记录//单个字段$sql = "select * from 表名 where 字段名 in "; $sql .= "(select 字段名 from 表名 group by 字段名 having count(字段名) > 1 )";//多个字段$sql = "select * from 表名 别名 where (别名.字段1,别名.字段2) in "; $sql .= "(select 字段1,字段2 from 表名 group by 字段1,字段2 having count(*) > 1 )";//删除表中多余的重复记录(留id最小)//单个字段$sql = "delete from 表名 where 字段名 in "; $sql .= "(select 字段名 from 表名 group by 字段名 having count(字段名) > 1) "; $sql .= "and 主键ID not in "; $sql .= "(select min(主键ID) from 表名 group by 字段名 having count(字段名 )>1) ";//多个字段$sql = "delete from 表名 别名 where (别名.字段1,别名.字段2) in "; $sql .= "(select 字段1,字段2 from 表名 group by 字段1,字段2 having count(*) > 1) "; $sql .= "and 主键ID not in "; $sql .= "(select min(主键ID) from 表名 group by 字段1,字段2 having count(*)>1) ";
业务篇
连续范围问题
//创建测试表CREATE TABLE `test_number` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT, `number` int(11) unsigned NOT NULL DEFAULT ‘0‘ COMMENT ‘数字‘, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8//创建测试数据insert into test_number values(1,1); insert into test_number values(2,2); insert into test_number values(3,3); insert into test_number values(4,5); insert into test_number values(5,7); insert into test_number values(6,8); insert into test_number values(7,10); insert into test_number values(8,11);
实验目标：求数字的连续范围。
根据上面的数据，应该得到的范围。
1-35-57-810-11//执行Sqlselect min(number) start_range,max(number) end_rangefrom( select number,rn,number-rn diff from ( select number,@number:[email protected]+1 rn from test_number,(select @number:=0) as number ) b ) c group by diff;
数字的连续范围
签到问题
//创建参考表(模拟数据需要用到)CREATE TABLE `test_nums` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT=‘参考表‘;//模拟数据，插入 1-10000 连续数据.//创建测试表CREATE TABLE `test_sign_history` ( `id` int(10) unsigned NOT NULL AUTO_INCREMENT, `uid` int(11) unsigned NOT NULL DEFAULT ‘0‘ COMMENT ‘用户ID‘, `create_time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP COMMENT ‘签到时间‘, PRIMARY KEY (`id`) ) ENGINE=InnoDB DEFAULT CHARSET=utf8 COMMENT=‘签到历史表‘;//创建测试数据insert into test_sign_history(uid,create_time) select ceil(rand()*10000),str_to_date(‘2016-12-11‘,‘%Y-%m-%d‘)+interval ceil(rand()*10000) minutefrom test_nums where id<31;//统计每天的每小时用户签到情况select h, sum(case when create_time=‘2016-12-11‘ then c else 0 end) 11Sign, sum(case when create_time=‘2016-12-12‘ then c else 0 end) 12Sign, sum(case when create_time=‘2016-12-13‘ then c else 0 end) 13Sign, sum(case when create_time=‘2016-12-14‘ then c else 0 end) 14Sign, sum(case when create_time=‘2016-12-15‘ then c else 0 end) 15Sign, sum(case when create_time=‘2016-12-16‘ then c else 0 end) 16Sign, sum(case when create_time=‘2016-12-17‘ then c else 0 end) 17Signfrom( select date_format(create_time,‘%Y-%m-%d‘) create_time, hour(create_time) h, count(*) c from test_sign_history group by date_format(create_time,‘%Y-%m-%d‘), hour(create_time) ) a group by h with rollup;
统计每天的每小时用户签到情况
//统计每天的每小时用户签到情况(当某个小时没有数据时，显示0)select h , sum(case when create_time=‘2016-12-11‘ then c else 0 end) 11Sign, sum(case when create_time=‘2016-12-12‘ then c else 0 end) 12Sign, sum(case when create_time=‘2016-12-13‘ then c else 0 end) 13Sign, sum(case when create_time=‘2016-12-14‘ then c else 0 end) 14Sign, sum(case when create_time=‘2016-12-15‘ then c else 0 end) 15Sign, sum(case when create_time=‘2016-12-16‘ then c else 0 end) 16Sign, sum(case when create_time=‘2016-12-17‘ then c else 0 end) 17Signfrom( select b.h h,c.create_time,c.c from ( select id-1 h from test_nums where id<=24 ) b left join ( select date_format(create_time,‘%Y-%m-%d‘) create_time, hour(create_time) h, count(*) c from test_sign_history group by date_format(create_time,‘%Y-%m-%d‘), hour(create_time) ) c on (b.h=c.h) ) a group by h with rollup;
统计每天的每小时用户签到情况(当某个小时没有数据时，显示0)
//统计每天的用户签到数据和每天的增量数据select type, sum(case when create_time=‘2016-12-11‘ then c else 0 end) 11Sign, sum(case when create_time=‘2016-12-12‘ then c else 0 end) 12Sign, sum(case when create_time=‘2016-12-13‘ then c else 0 end) 13Sign, sum(case when create_time=‘2016-12-14‘ then c else 0 end) 14Sign, sum(case when create_time=‘2016-12-15‘ then c else 0 end) 15Sign, sum(case when create_time=‘2016-12-16‘ then c else 0 end) 16Sign, sum(case when create_time=‘2016-12-17‘ then c else 0 end) 17Signfrom( select b.create_time,ifnull(b.c-c.c,0) c,‘Increment‘ type from ( select date_format(create_time,‘%Y-%m-%d‘) create_time, count(*) c from test_sign_history group by date_format(create_time,‘%Y-%m-%d‘) ) b left join ( select date_format(create_time,‘%Y-%m-%d‘) create_time, count(*) c from test_sign_history group by date_format(create_time,‘%Y-%m-%d‘) ) c on(b.create_time=c.create_time+ interval 1 day) union all select date_format(create_time,‘%Y-%m-%d‘) create_time, count(*) c, ‘Current‘ from test_sign_history group by date_format(create_time,‘%Y-%m-%d‘) ) a group by type order by case when type=‘Current‘ then 1 else 0 end desc;
统计每天的用户签到数据和每天的增量数据
//模拟不同的用户签到了不同的天数insert into test_sign_history(uid,create_time) select uid,create_time + interval ceil(rand()*10) day from test_sign_history,test_nums where test_nums.id <10 order by rand() limit 150;//统计签到天数相同的用户数量select sum(case when day=1 then cn else 0 end) 1Day, sum(case when day=2 then cn else 0 end) 2Day, sum(case when day=3 then cn else 0 end) 3Day, sum(case when day=4 then cn else 0 end) 4Day, sum(case when day=5 then cn else 0 end) 5Day, sum(case when day=6 then cn else 0 end) 6Day, sum(case when day=7 then cn else 0 end) 7Day, sum(case when day=8 then cn else 0 end) 8Day, sum(case when day=9 then cn else 0 end) 9Day, sum(case when day=10 then cn else 0 end) 10Dayfrom( select c day,count(*) cn from ( select uid,count(*) c from test_sign_history group by uid ) a group by c ) b;
统计签到天数相同的用户数量
//统计每个用户的连续签到时间select * from ( select d.*, @ggid := @cggid, @cggid := d.uid, if(@ggid = @cggid, @grank := @grank + 1, @grank := 1) grank from ( select uid,min(c.create_time) begin_date ,max(c.create_time) end_date,count(*) count from ( select b.*, @gid := @cgid, @cgid := b.uid, if(@gid = @cgid, @rank := @rank + 1, @rank := 1) rank, [email protected] flag from ( select distinct uid, date_format(create_time,‘%Y-%m-%d‘) create_time, datediff(create_time,now()) diff from test_sign_history order by uid,create_time ) b, (SELECT @gid := 1, @cgid := 1, @rank := 1) as a ) c group by uid,flag order by uid,count(*) desc ) d,(SELECT @ggid := 1, @cggid := 1, @grank := 1) as e )f where grank=1;
统计每个用户的连续签到时间
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Thanks ~