2015-03-19 80 views
-1

嗨,我是新开发者。请耐心等待我的编码。我创建了一个带有id(自动增量),状态,票证号码和代码的表格arlog。门票和代码编号被设置为唯一。即这个组合的副本不能再次插入。但个人票号或cpt可以插入多次。它工作正常。现在我想对于与另一个表使用select查询到arlog表票号和code.Here是select语句选择语句以显示2列组合的结果

$result = mysql_query("SELECT * FROM `ar` C WHERE provider='".$_SESSION['PROVIDER'] 
."' AND C.`TicketNo` IN (SELECT TicketNo FROM `arlog` L where L.status NOT IN('New','Completed','Completed(Ar_aging) 
','Completed(Rework)','Rework','Completed_Followup','Completed_Supervising' ) 
    and L.assign='".$_SESSION['NAME']."') order by id desc") or die(mysql_error()); 

查询检查arlog和显示票号正确地将。但我想要在arlog中结合TicketNo和Code。我做了研究但找不到解决方案。首先是可能的?

回答

0

请尝试以下SQL:

SELECT L.TicketNo ,L.Code,C.* FROM `ar` C left join `arlog` L ON C.TicketNo = L.TicketNo where C.provider='your condition' and L.status NOT IN('New','Completed','Completed(Ar_aging)','Completed(Rework)','Rework','Completed_Followup','Completed_Supervising') and L.assign='your condition' order by by C.id desc 

希望这能帮助你!