SELECT income, COUNT(*) AS cnt FROM Graduates GROUP BY income HAVING COUNT(*) >= ALL(SELECT COUNT(*) FROM Graduates GROUP BY income); 使用极值函数: SELECT income, COUNT(*) AS cnt FROM Graduates GROUP BY income HAVING COUNT(*) >= (SELECT MAX(cnt) FROM (SELECT COUNT(*) AS cnt FROM...
count(1) over(partition by docnum) count()函数统计字段docnum的数量,再分组字段docnum,不排序仅统计个数。 SELECT t.docnum FROM (select m.* , count(1) over (partition by docnum) as cnt from F0411 m) T WHERE T.STATUS = 'FULL' AND CNT = 1; 对T临时表进行过滤,让字段status等于full,...
,count(1) as cnt from (select t.user_id ,t.login_time ,date_sub(login_time, INTERVAL t.num DAY) as date_rslt from (select user_id ,login_time ,row_number() over(partition by user_id order by login_time) as num from login_log ) t ) a group by a.user_id ,a.date_rslt; ...
第一次,运行 count(*) SELECT COUNT(*) AS CNT FROM dbo.MobileLink 1. 2. 可以看到运行大约花了 3 秒时间 执行计划也简单,走了全表扫描 万能的性能杀-索引 我之前也分享过,数据是存在数据页上的。这个数据页可以看做是一页纸。在纸上把字写得越紧凑,得到的信息越多。反之,如果你把字...
SELECT count(*) AS cnt, package_name FROM ( SELECT t.a AS package_name FROM ( SELECT transform(packages_map_array, x -> Element_at(x, 'packageName')) AS package_array FROM ( SELECT cast(Json_extract(data_json, '$.packages') ...
SELECT COUNT (*) AS cnt, SUM (salary) AS sum_sal FROM employee2 WHERE deptno = 11 group by deptno; 1. 执行结果如图 清空表数据 truncate table TEST 1. 使用分析函数 sum (…) over (order by…) 可以生成累计和 使用分析函数 sum (…) over (order by…) 可以生成累计和,查询部门1下的工资...
SELECT 客户编号,COUNT(DISTINCT 产品代号) AS 变化次数 FROM TABLE GROUP BY 客户编号 便可以解决。如果会重复，oracle的写法是：select cino,sum(decode(pro_id,flag,0,1) )ch_cnt ( select cino,pro_id,lead(pro_id,1,pro_id)over(partition by pro_id order by 必须有排序字段)flag from...
方法1，建立一个模块 将工程的启动设置为 SUB MAIN 在模块里写下面代码 Public db As Database Public rs As Recordset Sub main()set db=db.opendatabase("数据库路径"<建议用相对路径--就是把数据库文件和工程文件放在同一目录下--相对路径就是APP.PATH+"数据库名带后缀">)第一启动窗体<比如...
count(*) over( ) cnt from test_youhua.test_avg_medium_freq ) as tmp --如果是奇数,取排序中间的,如果是偶数,取两个中间的均值 where if(cnt%2=0,num in(cnt/2,cnt/2+1),num=(cnt+1)/2)""").show() #数据量非常大时,这里或许可以直接使用num=ceil(cnt/2) ...
count(*) over(partition by LAT, LON) as cnt_2 FROM insurance ) a WHERE a.cnt_1 > 1 AND a.cnt_2 < 2 586. 订单最多的客户 难度简单 SQL架构 在表orders中找到订单数最多客户对应的customer_number。 数据保证订单数最多的顾客恰好只有一位。