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select*from(select date_add('2003-01-01 00:00:00.000',INTERVAL n5.num10000+n4.num1000+n3.num100+n2.num10+n1.num DAY)as date from(select 0 as num union all select 1 union all select 2 union all select ...

select a.Date from(select curdate()-INTERVAL(a.a+(10*b.a)+(100*c.a)+(1000*d.a))DAY as Date from(select 0 as a union all select 1 union all select 2 union all select 3 union all select 4 union all ...

var3:=SELECT[ALL|DISTINCT]var1.select_expr,var2.select_expr,. FROM@var1 join@var2 on.;INSERT OVERWRITE|INTO TABLE[PARTITION(partcol1=val1,partcol2=val2.)] SELECT[ALL|DISTINCT]select_expr,select_expr,....
Puppet 2022-04-03 22:30:39 0 浏览量 回答数 0

SELECT SUBSTRING_INDEX(SUBSTRING_INDEX(t.values,',',n.n),',',-1)value FROM table1 t CROSS JOIN(SELECT a.N+b.N*10+1 n FROM(SELECT 0 AS N UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ...

INTERVAL(b1.b+b2.b+b3.b+b4.b+b5.b+b6.b+b7.b+b8.b+b9.b+b10.b+b11.b+b12.b+b13.b+b14.b+b15.b+b16.b)DAY)AS ddate FROM(SELECT 0 AS b UNION SELECT 1)b1,(SELECT 0 AS b UNION SELECT 2)b2,(SELECT 0 AS b UNION ...

select company,avg(salary)as salary from select*from select*from select x.sort,x.company,x.salary,y.total_num from select row_number()over(partition by company order by salary)as sort,*from ...
o_coby 2022-04-03 19:24:12 0 浏览量 回答数 0

CREATE OR REPLACE VIEW generator_16 AS SELECT 0 n UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT ...

FORMAT(a.Date,"%b")as month,DATE_FORMAT(a.Date,"%m-%Y")as md,'0' as amount from(select curdate()-INTERVAL(a.a+(10*b.a)+(100*c.a))DAY as Date from(select 0 as a union all select 1 union all select 2 ...

SELECT TRANSFORM(key,value)USING 'cmd2' from SELECT TRANSFORM(*)USINg 'cmd1' from SELECt*FROM data distribute by col2 sort by col1 t distribute by key sort by value t2;官方文档说明见:...

select 2,'1' union all select 3,'4,3,2' union all select 4,'2,1' union all select 5,'1,4' union all select 6,'2,3,4' select count(distinct(value))as category from cte cross apply string_split(cte....

SELECT*FROM src2 WHERE value>0)b ON a.key=b.key c UNION ALL SELECT*FROM SELECT a.key,b.value FROM SELECT*FROM src WHERE key IS NOT NULL)a LEFT OUTER JOIN SELECT*FROM src3 WHERE value>0)b ON a.key=b....

4 as id2""").createOrReplaceTempView("table1")spark.sql("""select 1 as c1,'london' as city union select 4 as c1,'texas' as city""").createOrReplaceTempView("table2")spark.sql("""select 2 as c1,...

SELECT json_array_contains(‘[1,2,3]’,2);json_array_get(json_array,index)→varchar 将指定index处的元素返回到json数组中,index从0开始计数 SELECT json_array_get(‘[“a”,“b”,“c”]’,0);–‘a’SELECT ...
nicenelly 2022-04-07 06:26:56 2244 浏览量 回答数 0

SELECT t1.TBB,t2.ID FROM(SELECT DISTINCT tb.ID tbb FROM TESTB tb)t1,TESTA t2 WHERE NOT EXISTS(SELECT 1 FROM TESTB tb2 WHERE tb2.ID|tb2.ID1=t1.TBB|t2.ID)ORDER BY t1.TBB,t2.ID这样是可以的SELECT t1.TBB,t...

SELECT*FROM src2 WHERE value>0)b ON a.key=b.key c UNION ALL SELECT*FROM SELECT a.key,b.value FROM SELECT*FROM src WHERE key IS NOT NULL)a LEFT OUTER JOIN SELECT*FROM src3 WHERE value>0)b ON a.key=b....

select 'col2' union all select 'col3' union all select 'col4' b(column_name) 而不是： select ID,column_name,column_value From select ID,col1,col2,col3,col4 from table1 a unpivot column_value FOR ...

('#select1').select2({ allowClear:true, data:[{id:0,text:'hello'},{id:1,text:'world'}] });var str=\$('#select1').select2("data").text;结果这一步就是取不到值，网上查了一下，都是这么取值的，select2("val...
a123456678 2022-04-06 16:32:26 2277 浏览量 回答数 1

select project_id,employee_id from(select t1.project_id,t1.employee_id,t2.experience_years,rank()over(partition by t1.project_id order by t2.experience_years desc)as rank_num from tmp.table_project t1...

5）欧文22.181毫秒 SELECT s.stud_id,s.name FROM student s WHERE EXISTS(SELECT*FROM student_club WHERE stud_id=s.stud_id AND club_id=30)AND EXISTS(SELECT*FROM student_club WHERE stud_id=s.stud_id AND ...

SELECT column_name(s)FROM table2;注释：默认地，UNION 操作符选取不同的值。如果允许重复的值，请使用 UNION ALL。SQL UNION ALL 语法 SELECT column_name(s)FROM table1 UNION ALL SELECT column_name(s)FROM ...
Guardtime 2022-04-02 23:55:09 0 浏览量 回答数 0

select*from 表名 order by 列1 asc|desc[,列2 asc|desc,.] 如果列1的值相同，则按照列2排序，以此类推asc从小到大desc从大到小 例1.根据学院分组ID降序(desc) select CollegeID from StudentBindPaperTypeEntity ...

select t.a,t.b+t2.b,if(t.c=t2.c,t.c,1)from(SELECT a,b,c FROM q3 where b%2=1)t left join(select a,b,c from q3)t2 where t.a=t2.a and t.b+1=t2.b 5. rank(),dense_rank()select project_id,employee_id,dense_...

Code Load(0.6ms)SELECT `codes`.*FROM `codes` LIMIT 10 Code Load(0.9ms)SELECT `codes`.*FROM `codes` LIMIT 100Code Load(1.2ms)SELECT `codes`.*FROM `codes` LIMIT 200Code Load(1.4ms)SELECT `codes`....

insert into access_date select current_date-2,hll_add_agg(hll_hash_integer(user_id))from generate_series(9000,40000)t(user_id);postgres=select#userids from access_date where acc_date=current_date;...

CREATE VIEW generator_16 AS SELECT 0 n UNION ALL SELECT 1 UNION ALL SELECT 2 UNION ALL SELECT 3 UNION ALL SELECT 4 UNION ALL SELECT 5 UNION ALL SELECT 6 UNION ALL SELECT 7 UNION ALL SELECT 8 UNION ALL...

select 2 as NumberStart, 3 as NumberEnd union all select 5, 7 cte as select(select min(NumberStart)from Ranges)as rn, (select max(NumberEnd)from Ranges)as rn_end union all select rn+1, t.rn_end from ...

2.强制 Oracle 执行，方法是加上 BYPASS_UJVC 注释 UPDATE(SELECT a.user_name,b.user_name AS user_name01 FROM t_test01 a,t_test02 b WHERE a.user_id=b.user_id)SET user_name=user_name01;COMMIT;2、采用MERGE ...

c:=select*from pv1(src2,@another_day);d:=select*from@c union all select*from@b;with t as(select*from src3) select*from@c union all select*from@d union all select*from pv1(t,@another_day);
Puppet 2022-04-03 22:30:44 0 浏览量 回答数 0