我的表A如下:
ID | NegeriID | yeartamatkhidmat | yeartamatkhidmat
-----------------------------------------------------------------
1 | 2 | 2001 | 2002
1 | 2 | 2003 | 2007
2 | 2 | 2008 | 2012
2 | 1 | 2013 | 2018
3 | 3 | 2000 | 2001
根据Negeri(州),我有一个子查询来获取多年工作经验的总和:
$query = DB::table('B')
->select('B.name','B.ID', '');
$query->select(DB::raw('B.name, m.NegeriID, sum(distinct m.duration)'))
->from(DB::raw('(SELECT ID, NegeriID, yeartamatkhidmat - yearmulakhidmat as duration FROM A) AS m
RIGHT JOIN B ON B.ID=m.ID'))
->groupBy('m.ID', 'm.NegeriID');
我还有另一个表,其名称和值为Negeri:
表B
ID | Name
--------------
1 | Bill
2 | Jane
3 | Roy
表C
NegeriID | Negeri
--------------
1 | E
2 | F
3 | G
如何从表中获取name和Negeri的值以集成到上面的子查询中?
我尝试过以下代码:
$query->JOIN(DB::RAW('(SELECT m.itemRegistrationID, m.NegeriID, sum(distinct m.duration) AS s
FROM (SELECT itemRegistrationID, NegeriID, yeartamatkhidmat - yearmulakhidmat as duration FROM itemregistrationpangkat) AS m)'),
function($join){
$join->on("itemregistrations.itemRegistrationID","=","m.itemRegistrationID");
})
->groupBy('m.itemRegistrationID', 'm.NegeriID');
但显示错误:
Syntax error or access violation: 1248 Every derived table must have its own alias