求教 mysql 查询语句。mysql 数据格式如下，表中 app 重复。需求：取出一段时间内（四天内吧）每天每个 appid 调用总量（ count 总量）最大的前两个 appid

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This topic created in 3989 days ago, the information mentioned may be changed or developed.

logtime appid count
100018 1226461
1002495 6165611
2015-10-1 1000255 156164611
100018 4964610
1002495 16461661
1000255 46461168

100018      196494561
            1002495     416461616
            1000255     616113

2015-10-2 100018 16461131
1002495 16461613
1000255 1641351

100018      2646116
            1002495     6116161
            1000255     126461611

2015-10-3 100018 548531
1002495 54136116
1000255 496811

100018      41646161
            1002495     1644131
            1000255     61164611

2015-10-4 100018 443134
1002495 15133
1000255 1641434

27 replies • 2015-10-27 08:14:30 +08:00

luefei

Oct 26, 2015

。。。这个格式。。好蛋疼。。

luefei

Oct 26, 2015

logtime appid count
100018 1226461
1002495 6165611
2015-10-1 1000255 156164611
100018 4964610
1002495 16461661
1000255 46461168

100018 196494561
1002495 416461616
1000255 616113
2015-10-2 100018 16461131
1002495 16461613
1000255 1641351

100018 2646116
1002495 6116161
1000255 126461611
2015-10-3 100018 548531
1002495 54136116
1000255 496811

100018 41646161
1002495 1644131
1000255 61164611
2015-10-4 100018 443134
1002495 15133
1000255 1641434

luefei

Oct 26, 2015

herojaxy

Oct 26, 2015

这排版...没看懂啥意思

luefei

Oct 26, 2015

@herojaxy 额数据格式三列日期 appid 调用量，结果发出去后日期那一列就合进去了。第一次发问。不知道怎么发图片。尝试了两次这个格式都没调整对。

bugsnail

Oct 26, 2015

表名，字段名，必须
数据一行就可以了，只是为了形式
关开格式的问题，不单单只能用空格的，你可以把空格替换成--再发上来都可以.....

最后，自行 google V2EX 发图 ....

HanSonJ

Oct 26, 2015

select count(appid) from table where time > 4days group by appid limit 2 order by count(appid) desc
这样取出最大两个的总数,大概就这意思，不知道符不符合

luefei

Oct 26, 2015

@bugsnail 谢谢。我重新整理下

luefei

Oct 26, 2015

@HanSonJ 这样是没有调用总量的。调用总量还要求和。

luefei

Oct 26, 2015

@bugsnail
logtime appid count
------------100018 1226461
------------1002495 6165611
2015-10-1 1000255 156164611
100018 4964610
1002495 16461661
1000255 46461168

100018 196494561
1002495 416461616
1000255 616113
2015-10-2 100018 16461131
1002495 16461613
1000255 1641351

100018 2646116
1002495 6116161
1000255 126461611
2015-10-3 100018 548531
1002495 54136116
1000255 496811

100018 41646161
1002495 1644131
1000255 61164611
2015-10-4 100018 443134
1002495 15133
1000255 1641434

luefei

Oct 26, 2015

logtime appid count
------------ ------100018 1226461
-------------------1002495 6165611
2015-10-1 1000255 156164611
-------------------100018 4964610
-------------------1002495 16461661
-------------------1000255 46461168

-------------------100018 196494561
-------------------1002495 416461616
-------------------1000255 616113
2015-10-2 100018 16461131
-------------------1002495 16461613
-------------------1000255 1641351

-------------------100018 2646116
-------------------1002495 6116161
-------------------1000255 126461611
2015-10-3 100018 548531
-------------------1002495 54136116
-------------------1000255 496811

-------------------100018 41646161
-------------------1002495 1644131
-------------------1000255 61164611
2015-10-4 100018 443134
-------------------1002495 15133
-------------------1000255 1641434

popok

Oct 26, 2015

@luefei 你还是直接图片随便上传一个地方，贴图片地址吧

iyaozhen

Oct 26, 2015

取 4 天？每天的调用量最大的 2 个？

popok

Oct 26, 2015

select sum(count) from 表名 group by appid where logtime>2015-10-20 and logtime<2015-10-24 limit 0,2

我猜是这样，没测试过

popok

Oct 26, 2015

select sum(count) from 表名 where logtime>2015-10-20 and logtime<2015-10-24 group by appid order by sum(count) desc limit 0,2

luefei

Oct 26, 2015

@popok 恩恩。我看下重发个问题吧。这个第一次提问，搞砸了。。

popok

Oct 26, 2015

@luefei 你的问题描述其实很简单，直接一个表头就说明白了，不需要截图了

popok

Oct 26, 2015

select appid,sum(count) from 表名 where logtime>2015-10-20 and logtime<2015-10-24 group by appid order by 2 desc limit 0,2

这个应该就是结果了

bugsnail

Oct 26, 2015

结构不知道是不是这样...
![]( http://7xiwkv.com1.z0.glb.clouddn.com/v2test.png)

sql:

```
select login_date,appid,count
from t_name
where UNIX_TIMESTAMP('2015-10-22') BETWEEN UNIX_TIMESTAMP(CURDATE())-3600*24*4 and UNIX_TIMESTAMP(CURDATE())
group by login_date,appid
order by login_date,count desc
```
那个 22 号改为当前日期可以查看过去 4 天的，那个分组前两条不会了，请大神....

luefei

Oct 26, 2015

@popok 恩恩。算是比较复杂了。我写出来的语句太复杂效率太低用不了。。所以还是请教帮忙，技术水平有限。 select a.* from (select logTime, appid, sum(count) as dailycount from table where datediff(current_date(), logTime) < 4 group by logTime, appid order by logTime, dailycount desc) as a where 2 > (select count(*) from (select logTime, appid, sum(count) as b from table where datediff(current_date(), logTime) group by logTime, appid order by logTime, dailycount desc) as b where a.logTime = b.logTime and b.dailycount > a.dailycount);

popok

Oct 26, 2015

@luefei 好吧，你这个我看的晕，然后重新看了下主题，好像我理解错了

tusj

Oct 26, 2015

DROP TABLE IF EXISTS test.app_usage;
CREATE TABLE test.app_usage
(
log_time DATETIME,
app_id BIGINT,
log_count BIGINT
);

DELETE FROM test.app_usage;
INSERT INTO test.app_usage VALUES (NOW(), 123, 100);
INSERT INTO test.app_usage VALUES (NOW(), 456, 200);
INSERT INTO test.app_usage VALUES (NOW(), 789, 300);
INSERT INTO test.app_usage VALUES (NOW(), 124, 100);
INSERT INTO test.app_usage VALUES (NOW(), 234, 200);
COMMIT;

-- SELECT * FROM test.app_usage;
SELECT a.*
FROM (
SELECT concat(DATE_FORMAT(log_time, '%Y%m%d'), app_id) AS group_key, MAX(date_format(log_time, '%Y%m%d')) as log_date, MAX(app_id), SUM(log_count) AS daily_count
FROM test.app_usage
WHERE DATEDIFF(CURRENT_DATE(), log_time) < 4
GROUP BY group_key) AS a
ORDER BY a.daily_count DESC
LIMIT 0, 1;

-- DROP TABLE IF EXISTS test.app_usage;

luefei

Oct 26, 2015

@popok 恩恩。需求略复杂。。智商捉急所以请求大牛们帮忙解决。

luefei

Oct 26, 2015

@bugsnail 谢谢指导

luefei

Oct 26, 2015

@tusj 谢谢。这个语句里的 sum(log_count) 是四天的调用量和，不是每天的 appid 求和。

zeayes

Oct 26, 2015

USE test;
DROP TABLE `test`;
CREATE TABLE `test`(
`logtime` DATETIME NOT NULL,
`appid` VARCHAR(12) NOT NULL,
`count` INT NOT NULL DEFAULT 0
)DEFAULT CHARSET = UTF8;

INSERT INTO `test` VALUES ('2015-10-15', '10001', 100);
INSERT INTO `test` VALUES ('2015-10-15', '10001', 150);
INSERT INTO `test` VALUES ('2015-10-15', '10001', 50);
INSERT INTO `test` VALUES ('2015-10-15', '10003', 70);
INSERT INTO `test` VALUES ('2015-10-15', '10002', 420);
INSERT INTO `test` VALUES ('2015-10-15', '10002', 120);
INSERT INTO `test` VALUES ('2015-10-15', '10002', 190);
INSERT INTO `test` VALUES ('2015-10-16', '10001', 110);
INSERT INTO `test` VALUES ('2015-10-16', '10001', 130);
INSERT INTO `test` VALUES ('2015-10-16', '10001', 90);
INSERT INTO `test` VALUES ('2015-10-15', '10002', 200);
INSERT INTO `test` VALUES ('2015-10-16', '10003', 40);
INSERT INTO `test` VALUES ('2015-10-16', '10003', 80);
INSERT INTO `test` VALUES ('2015-10-16', '10003', 60);

set @num := 0, @logtime := '';
select logtime, appid, dailycount from (select a.logtime, a.appid, a.dailycount, @num := if(@logtime = date(a.logtime), @num + 1, 1) as row_number, @logtime := a.logtime from (select logtime, appid, sum(count) as dailycount from test group by logtime, appid order by logtime desc, dailycount desc) as a) as b where b.row_number < 3;

http://www.xaprb.com/blog/2006/12/07/how-to-select-the-firstleastmax-row-per-group-in-sql/

tusj

Oct 27, 2015 via Android

@luefei 我给的就是按每天，每个 app_id 汇总的啊
不然你觉得 group by 拿来干什么吃的呢？
可能是你被 limit0,1 迷惑了，你把 limit 去掉就看明白了

求教 mysql 查询语句。mysql 数据格式如下，表中 app 重复。需求：取出 一段时间内（四天内吧） 每天每个 appid 调用总量（ count 总量）最大的前两个 appid

求教 mysql 查询语句。mysql 数据格式如下，表中 app 重复。需求：取出一段时间内（四天内吧）每天每个 appid 调用总量（ count 总量）最大的前两个 appid