求一条SQL语句

2010-06-10 04:47:32 168 阅读 AI 写同款 · GEO 优化AI 选题数据监测一键分发 ›

163种中草药图像数据集构建与ResNet-50迁移学习实战指南在计算机视觉领域，高质量标注数据集是模型训练的基石，直接影响图像分类、目标检测等任务的性能。其核心原理在于为算法提供结构化的监督信号，通过深度学习模型学习从像素到语义的映射关系。构建此类数据集的技术价值在于解决了特定垂直领域（如中医药）数据稀缺的痛点，为AI技术落地提供了关键“燃料”。在应用场景上，标准化数据集广泛支撑着教学演示、算法验证和产业应用开发。本文聚焦于中草药识别这一细分领域，详细介绍了一个包含163种常见药材、经过多源聚合与清洗的图像数据集构建全过程，并提供了基于PyTorch和ResNet-5 阅读详情

表A
id name bankaccount
100 aaa 123000
200 bbb 321000

表B
id amount
100 200
100 310
200 450
200 540
200 600

期望得到的结果是
id name bankaccount amount
100 aaa 123000 510
200 bbb 321000 1590

...全文

IEPE加速度同步采集卡、振动信号同步模块、温震一体八路高速采集 WJ288产品是一种物联网和工业以太网采集模块，实现了传感器数据采集并发送到网络。IEPE是“Integrated Electronics Piezo-Electric”的缩写，即集成电路型压电式传感器。这种带内置电子器件的压电传感器已成为工业应用领域的标准。半导体电路将压电传感器的高阻抗信号转换为更容易传输的低阻抗电压信号。借助IEPE，只需一根任意长度的标准电缆即可实现电源供给和信号传输，不会因电缆移动而产生测量误差。阅读详情

whiteseal

MS-SQL Server 34,876 成员 254,635 内容

+ 关注 + 加入社区

切换为时间正序

极夜之辉 2010-06-11

打赏
举报



select 

a.id,a.name,a.bankaccount,sum(b.amount) amount

from 

a left join b

on

a.id=b.id

lucky_2005 2010-06-11

打赏
举报

select a.id,a.name,a.bankaccount,sum(b.amount)
from a,b
where a.id = b.id
group by a.id,a.name,a.bankaccount

xiaoxiao8372 2010-06-11

打赏
举报

写地都挺好啊，我就不献丑了啊，哈哈！！

zhengtw 2010-06-11

打赏
举报

select A.id, A.name, A.bankaccount, sum(B.amount) as amount

from A, B

where A.id = B.id

group by A.id, A.name, A.bankaccount

aaajedll 2010-06-11

打赏
举报

b表GROUP BY ,SUM，然后和A表内连接就好了

宇峰科技 2010-06-11

打赏
举报

这里的SQL 高手回答问题好快的啊

心中的彩虹 2010-06-11

打赏
举报



if object_id('a') is not null drop table a

go

create table a(id int,name varchar(3),bankaccount int)

insert a

select 100,'aaa',123000 union all

select 200,'bbb',321000



if object_id('b') is not null drop table b

go

create table b(id int,amount int)

insert b

select 100,200 union all

select 100,310 union all

select 200,450 union all

select 200,540 union all

select 200,600



select a.id,bankaccount,sum(amount) amount

from a,b

where a.id=b.id

group by a.id,bankaccount

--结果

id      bankaccount     amount

100	123000	        510

200	321000	        1590

claro 2010-06-11

打赏
举报

帮顶回复内容太短了！

lxy_swufe 2010-06-10

打赏
举报

select id，name， bankaccount ，amount
from table_A,(select table_B.id,sum(amount) from table_B group by table_B.id) table_C
where table_A.id = table_C.id;

htl258_Tony 2010-06-10

打赏
举报

--> 生成测试数据表: [A]

IF OBJECT_ID('[A]') IS NOT NULL

	DROP TABLE [A]

GO

CREATE TABLE [A] ([id] [int],[name] [nvarchar](10),[bankaccount] [int])

INSERT INTO [A]

SELECT '100','aaa','123000' UNION ALL

SELECT '200','bbb','321000'



--> 生成测试数据表: [B]

IF OBJECT_ID('[B]') IS NOT NULL

	DROP TABLE [B]

GO

CREATE TABLE [B] ([id] [int],[amount] [int])

INSERT INTO [B]

SELECT '100','200' UNION ALL

SELECT '100','310' UNION ALL

SELECT '200','450' UNION ALL

SELECT '200','540' UNION ALL

SELECT '200','600'



--SELECT * FROM [A]

--SELECT * FROM [B]



-->SQL查询如下:



SELECT *

FROM a 

	CROSS APPLY(

		SELECT SUM(amount)amount 

		FROM b

		WHERE a.id=b.id

	) c

/*

id          name       bankaccount amount

----------- ---------- ----------- -----------

100         aaa        123000      510

200         bbb        321000      1590



(2 行受影响)

*/

2005

ChinaITOldMan 2010-06-10

打赏
举报

select a.id,a.name,a.bankaccount,sum(b.amount) as amount
from a,b
where a.id=b.id
group by a.id,a.name,a.bankaccount

andy_liucj 2010-06-10

打赏
举报

select a.id,a.name,a.bankaccount, t.amount from a join (

select id,sum(amount) amount from b group by id)t on t.id=a.id

wqmgxj 2010-06-10

打赏
举报

select a.*,c.amount

from a, (selecvt id,sum(amount) as amount from b group by id) c

where a.id=c.id

wqmgxj 2010-06-10

打赏
举报

select a.*,c.amount

from a, (selecvt id,sum(amount) as amount from b group by id) c

where a.id=c.id

cxmcxm 2010-06-10

打赏
举报

select a.*,c.amount

from a, (selecvt id,sum(amount) as amount from b group by id) c

where a.id=c.id

dawugui 2010-06-10

打赏
举报

select a.* , sum(b.amount) amount from a , b where a.id = b.id group by a.id , a.name , a.bankaccount



select a.* , (select sum(b.amount) from b where b.id = a.id) amount from a

wujinyuan 2010-06-10

打赏
举报

select a.*,t.amount from a left join (select b.ID,amount=sum(b.amount) from b group by b.ID) t on a.ID=b.id

YangWenChaoX 2010-06-10

打赏
举报

[Quote=引用 3 楼 josy 的回复:]

SQL code
---测试数据---
if object_id('[A]') is not null drop table [A]
go
create table [A]([id] int,[name] varchar(3),[bankaccount] int)
insert [A]
select 100,'aaa',123000 union all
select 200,'bbb',321……
[/Quote]
甚好！每次都提供测试数据。。。

百年树人 2010-06-10

打赏
举报

---测试数据---

if object_id('[A]') is not null drop table [A]

go

create table [A]([id] int,[name] varchar(3),[bankaccount] int)

insert [A]

select 100,'aaa',123000 union all

select 200,'bbb',321000

if object_id('[B]') is not null drop table [B]

go

create table [B]([id] int,[amount] int)

insert [B]

select 100,200 union all

select 100,310 union all

select 200,450 union all

select 200,540 union all

select 200,600

 

---查询---

select a.id,a.name,a.bankaccount,sum(b.amount) as amount

from a,b

where a.id=b.id

group by a.id,a.name,a.bankaccount



---结果---

id          name bankaccount amount

----------- ---- ----------- -----------

100         aaa  123000      510

200         bbb  321000      1590



(2 行受影响)

永生天地 2010-06-10

打赏
举报



select 

a.id,a.name,a.bankaccount,sum(b.amount) amount

from 表A a left join 表B b on a.id=b.id group by a.id

加载更多回复（1）

求一条SQL语句

相关推荐