大致意思是:select stock_no,packageno,lotno,count(lotno) from `stock_detail` left join `kw_manage` on `stock_detail`.cust_no = `kw_manage`.cust_no left join `barcode_2d` on `kw_manage`.packageno = `barcode_2d`.packageno
是不是我的表达有问题?其实我想要的结果就是select stock_no,cust_no from `stock_detail` order by arr_time,adress,c_type,select rkno,packageno from `kw_manage` order by rkno asc,select packageno,lotno,count(lotno) from `barcode_2d` group by lotno order by lotno,packageno asc的结果联合查询。显示字段stock_no,rkno,packageno,lotno的关系即可。
select replace(mid('XABEA01-120NA12X1501120N310F6 1VA5A0023',26,11),' ','') 是例子,自行修改你的SQL语句有问题? select stock_no,cust_no from `stock_detail` order by arr_time,adress,c_type,select rkno,packageno from `kw_manage` order by rkno asc,select packageno,lotno,count(lotno) from `barcode_2d` group by lotno order by lotno,packageno asc
(3, 'SU13080800340', 'B', '2013-08-14 09:00:00', 'P32E', 'K33I196', '310F61VA5A', 1, 16, '2013-08-10 15:00:00', '2013-08-10 15:30:00'), (4, 'SU13080800340', 'B', '2013-08-14 09:00:00', 'P32E', 'K33I195', '310F61VA5A', 2, 16, '2013-08-10 15:00:00', '2013-08-10 15:30:00'),蓝色部分代表stock_no分别对应1个和2个packageno,也就是说例中这个stock_no一共对应3个packageno。所以说最重要的部分就是用select rkno,packageno from `kw_manage` order by rkno asc的结果来对应select stock_no,cust_no from `stock_detail` order by arr_time,adress,c_type asc的结果.排在最前的3个packageno对应stock_no的结果。(例子中是2个)
大致意思是:select stock_no,packageno,lotno,count(lotno) from `stock_detail` left join `kw_manage` on `stock_detail`.cust_no = `kw_manage`.cust_no left join `barcode_2d` on `kw_manage`.packageno = `barcode_2d`.packageno
格式是否固定,用MID、LEFTG、RIGHT等等字符串函数即可第1个问题,要求结果是什么
表`stock_detail`按照arr_time,adress,c_type升序排序。
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015001 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
SU13080800340 201210180001 CVT121015002 2X15
是例子,自行修改你的SQL语句有问题?
select stock_no,cust_no from `stock_detail` order by arr_time,adress,c_type,select rkno,packageno from `kw_manage` order by rkno asc,select packageno,lotno,count(lotno) from `barcode_2d` group by lotno order by lotno,packageno asc
(4, 'SU13080800340', 'B', '2013-08-14 09:00:00', 'P32E', 'K33I195', '310F61VA5A', 2, 16, '2013-08-10 15:00:00', '2013-08-10 15:30:00'),蓝色部分代表stock_no分别对应1个和2个packageno,也就是说例中这个stock_no一共对应3个packageno。所以说最重要的部分就是用select rkno,packageno from `kw_manage` order by rkno asc的结果来对应select stock_no,cust_no from `stock_detail` order by arr_time,adress,c_type asc的结果.排在最前的3个packageno对应stock_no的结果。(例子中是2个)
第一个sql语句少了asc
查询结果如下:INSERT INTO `stock_detail` (`stock_no`, `cust_no`) VALUES
('SU13080800340', '310F61VA5A'),
('SU13080800340', '310F61VA5A'),
('SU13080800345', '310F61VA5A'),
('SU13080800345', '310F61VA5A'),
('SU13080800345', '310F61VA5A'),
('SU13080800338', '310F61VA5A'),
('SU13080800338', '310F61VA5A'),
('SU13080800346', '310F61VA5A'),
('SU13080800346', '310F61VA5A');第二个sql语句缺少了cust_no字段,查询结果:INSERT INTO `kw_manage` (`rkno`, `packageno`, `cust_no`) VALUES
('201210180001', 'CVT121015001', '310F6 1VA5A'),
('201210180001', 'CVT121015002', '310F6 1VA5A');又已知select stock_no,cust_no,mount from `stock_detail`where stock_no = 'SU13080800340'的结果是:INSERT INTO `stock_detail` (`stock_no`, `cust_no`, `mount`) VALUES
('SU13080800340', '310F61VA5A', 1),
('SU13080800340', '310F61VA5A', 2);其中字段mount代表每个stock_no对应的packageno的数量,也就是说例中的stock_no一共对应3个packageno。
我需要求的核心部分就是用select rkno,packageno,cust_no from `kw_manage` order by rkno asc的结果来对应select stock_no,cust_no from `stock_detail` order by arr_time,adress,c_type asc的结果。例中是取这2个packageno的结果和stock_no对应(不足3个)
和9#类似,应该是这样的。//如果存在其他的符合条件的packageno,结果应该是这样的
stock_no rkno packageno
SU13080800340 201210180001 CVT121015001
SU13080800340 201210180001 CVT121015002
SU13080800340 201210180001 CVT121015002
SU13080800345 201210180002 ...
SU13080800345 201210180002 ...
...
不好意思,刚才网络断了
希望得到的测试结果://依据顶楼的数据,结果应该是这样的(kw_manage和stock_detail表联合查询的结果)
stock_no rkno packageno
SU13080800340 201210180001 CVT121015001
SU13080800340 201210180001 CVT121015002