Question

 SELECT t.Padal, u.Nool 
   FROM  (SELECT m.*, @n1:=@n1 + 1 num, 
                 @n2:=IF(Padal = 'பசிப் பிடி உதைத்த ஓமைச் செவ் வரை' , @
                         n1, @n2) pos 
            FROM tbl_padal m, (SELECT @n1:=0, @n2:=0) n 
           ORDER BY m.uqPadalid 
         ) t, tbl_Nool u 
  WHERE t.uqnoolId = u.uqnoolID 
    and u.nool = 'நற்றிணை' 
    and @n2 >= num - 2 
    AND @n2 <= num + 2

我想要获得两个记录（总共五个记录）的上方和下方

Answer 1

@mcbalaji 对于大纲，让我们从简单和已知的步骤逐步构建sql。创建并运行其中的每一个 -

1st  SELECT m.* From tbl_padal  m    (note that it is not ordered by uqPadalid)

2nd  SELECT m.* From tbl_padal  m  ORDER By uqPadalid     (now it is ordered)

3rd  SELECT 
     (SELECT Count(*) + 1 FROM tbl_padal AS A WHERE A.uqPadalid < m.uqPadalid ) as RowNum,
     m.*  From tbl_padal  m  ORDER By m.uqPadalid         (now it also has RowNum-s)

4th  So lets pick the central one that we want--
     SELECT RowNum FROM  
       (   SELECT 
           (SELECT Count(*) + 1 FROM tbl_padal AS A WHERE A.uqPadalid < m.uqPadalid ) as RowNum,
           m.Padal,
           m.uqPadalid  From tbl_padal  m  ORDER By m.uqPadalid        
       )
     Where m.Padal = 'your constant' 



Lastly--

SELECT  
  nums.RowNum,
  mm.RowNum,
  mm.Padal,
  mm.uqPadalid
  u.nool

FROM  (
  (   SELECT RowNum, uqPadalid FROM  
       (   SELECT 
           (SELECT Count(*) + 1 FROM tbl_padal AS A WHERE A.uqPadalid < m.uqPadalid ) as RowNum,
           m.Padal,
           m.uqPadalid  From tbl_padal  m  ORDER By m.uqPadalid        
       )
     Where m.Padal = 'constant1' 
  ) as nums

LEFT JOIN

  (   SELECT 
           (SELECT Count(*) + 1 FROM tbl_padal AS A WHERE A.uqPadalid < m.uqPadalid ) as RowNum,
           m.Padal,
           m.uqPadalid  From tbl_padal  m  ORDER By m.uqPadalid  
  ) as mm
   ON                      mm.RowNum  <=   nums.RowNum +2 
    AND nums.Rownum -2 <= mm.RowNum )

LEFT JOIN
   tbl_nool  u
Where mm.uqPadalid = u.uqPadalid
And  u.nool = 'constant2'

Order By  mm.RowNum

MS ACCESS查询表达式中的语法错误（缺少运算符）

1 个答案: