您的查询是正确的。你可能想看看你是如何填充底层表中的数据的。如果您使用SELECT语句作为在某些GUI中直接编辑的表格,可能会对如何创建2个不同的用户行来存储数据感到困惑。
反正我重构你的例子在MySQL中,填充2个表如下:
insert into `users` (name, phone) values ('Borrower01', '123456789');
insert into `users` (name, phone) values ('Borrower02', '234567890');
insert into `users` (name, phone) values ('Borrower03', '345678901');
insert into `users` (name, phone) values ('Borrower04', '456789012');
insert into `users` (name, phone) values ('Lender03', '123456789');
insert into `users` (name, phone) values ('Lender04', '234567890');
insert into `users` (name, phone) values ('Lender01', '345678901');
insert into `users` (name, phone) values ('Lender02', '456789012');
insert into loans (amt,date,pay_period,borrower_id,lender_id)
values (100.00, '2013-04-01', '2013-04-15',1,7)
insert into loans (amt,date,pay_period,borrower_id,lender_id)
values (100.00, '2013-04-01', '2013-04-15',2,8)
insert into loans (amt,date,pay_period,borrower_id,lender_id)
values (100.00, '2013-04-01', '2013-04-15',3,5)
insert into loans (amt,date,pay_period,borrower_id,lender_id)
values (100.00, '2013-04-01', '2013-04-15',4,6)
因此,贷款1和3应该由你的查询返回。
Select
loans.amt,
loans.date,
loans.pay_period,
borrower.phone As borrower_phone,
borrower.name As borrower_name,
lender.phone As lender_phone,
lender.name As lender_name
From
loans Left Join
users borrower On borrower.id = loans.borrower_id Left Join
users lender On lender.id = loans.lender_id
Where
lender.phone = '123456789'
or
borrower.phone = '123456789'
这回:
amt, date, pay_period, borrower_phone, borrower_name, lender_phone, lender_name
100, 2013-04-01, 2013-04-15, 123456789, Borrower01, 345678901, Lender01
100, 2013-04-01, 2013-04-15, 345678901, Borrower03, 123456789, Lender03
所以,查询是否已正确形成。检查你的数据。
它看起来像你修剪了表格信息,所以我们不必筛选一堆额外的信息(谢谢),但是你引用了一些不在你的表格中的字段。既然你没有收到错误,我猜'loan.borrower_id','loans.lender_id'和'users.id'确实存在但是没有列出? – Windle
我已添加它们。 – Smith
我的建议是发布每个表的一些示例数据以及查询的所需结果。 – Taryn