2015-03-19 113 views
-1

嗨,我是新開發者。請耐心等待我的編碼。我創建了一個帶有id(自動增量),狀態,票證號碼和代碼的表格arlog。門票和代碼編號被設置爲唯一。即這個組合的副本不能再次插入。但個人票號或cpt可以插入多次。它工作正常。現在我想對於與另一個表使用select查詢到arlog表票號和code.Here是select語句選擇語句以顯示2列組合的結果

$result = mysql_query("SELECT * FROM `ar` C WHERE provider='".$_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()); 

查詢檢查arlog和顯示票號正確地將。但我想要在arlog中結合TicketNo和Code。我做了研究但找不到解決方案。首先是可能的?

回答

0

請嘗試以下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 

希望這能幫助你!