我有两张表account_agent
和bulk_index
。每个表的数据如下
account_agent
name code account remark scheme
avi 10 151362 first as
babi 11 123478 new as
avi 10 151721 new as
avi 10 151765 new as
sam sas/01 925458 tin sas
bulk_index
agentcode account customer_name
10 151362 Alok
22 265721 Rohan
10 151721 Akash
使用这些表我想在account_agent
表上得到如下结果
account_agent
name code account remark
avi 10 151362 first
avi 10 151721 new
我已经尝试了查询
select * from account_agent where code = '10' and account = (select account from bulk_index where code = '10')
答案 0 :(得分:2)
这个查询对我有用
SELECT A.name, A.code, A.Account, A.remark, A.scheme
FROM account_agent A
INNER JOIN bulk_index B ON A.account = B.account
WHERE A.scheme = 'as' AND A.code = '10'
答案 1 :(得分:0)
这将只返回bulk_index表中相关的数据:
SELECT A.name, A.code, A.Account, A.remark
FROM account_agent A
INNER JOIN bulk_index B
ON A.account = B.account AND A.code = B.agentcode
答案 2 :(得分:0)
假设您的account
列是唯一的,您可以只执行一个简单的连接和一个where子句来获取此内容:
SELECT A.name, A.code, A.Account, A.remark
FROM account_agent A
INNER JOIN bulk_index B ON A.account = B.account
WHERE a.scheme like '%as%' and a.code = '10'
--remove the preceeding or succeeding % in like as needed
如果您只对scheme
列等于'as'并且code
等于10的值感兴趣,那么您可以这样做:
SELECT A.name, A.code, A.Account, A.remark
FROM account_agent A
INNER JOIN bulk_index B ON A.account = B.account
WHERE a.scheme = 'as' and a.code = '10'
以下是W3Schools页面的链接,该页面解释了在sql中使用like运算符。
答案 3 :(得分:0)
select name,code,account,remark
from account_agent
where exists(select bulk_index.agentcode from bulk_index where bulk_index.agentcode = account_agent.code)
此代码将显示bulk_index表格中存在的account_agent表代码的所有数据 agentcode