获得一对 table 的所有 ID 的最大值

Question

我是 sql 的初学者，我想知道如何为两个连接表的每个 id 获取几个整数的最大值？

到目前为止我做了什么：

SELECT 
  Reduced_T_Rappro_N.[Rente RES] AS Rente_RES, 
  Reduced_T_Rappro_N.[Montant capital constitutif] AS Montant_Capital_Consitutif_N, 
  [Reduced_T_Rappro_N - 1].[Montant capital constitutif] AS [ "Montant_Capital_Constitutif_N-1" ], 
  IIf(
    [Reduced_T_Rappro_N - 1].[Montant capital constitutif] = 0, 
    0, 
    (
      [Reduced_T_Rappro_N].[Montant capital constitutif] - [Reduced_T_Rappro_N - 1].[Montant capital constitutif]
    )/ [Reduced_T_Rappro_N - 1].[Montant capital constitutif]
  )* 100 AS [ "Evolution capital constitutif" ], 
  [Reduced_T_Rappro_N].[Montant Sous Rente], 
  [Reduced_T_Rappro_N - 1].[Montant Sous Rente], 
  (
    IIf(
      [Reduced_T_Rappro_N - 1].[Montant Sous Rente] = 0, 
      0, 
      (
        [Reduced_T_Rappro_N].[Montant Sous Rente] - [Reduced_T_Rappro_N - 1].[Montant Sous Rente]
      )/ [Reduced_T_Rappro_N - 1].[Montant Sous Rente]
    )
  )* 100 AS [ "Evolution montant sous rente" ], 
  Reduced_T_Rappro_N.[Montant rente initial] AS Montant_Rente_Initial_N, 
  [Reduced_T_Rappro_N - 1].[Montant rente initial] AS [ "Montant_Rente_Initial_N-1" ], 
  (
    IIf(
      [Reduced_T_Rappro_N - 1].[Montant rente initial] = 0, 
      0, 
      (
        [Reduced_T_Rappro_N].[Montant rente initial] - [Reduced_T_Rappro_N - 1].[Montant rente initial]
      )/ [Reduced_T_Rappro_N - 1].[Montant rente initial]
    )
  )* 100 AS [ "Evolution rente initial" ], 
  MAX(
    [Reduced_T_Rappro_N - 1].[Montant Sous Rente] + [Reduced_T_Rappro_N].[Montant Sous Rente]
  ) AS [ "Addition" ] 
FROM 
  Reduced_T_Rappro_N 
  INNER JOIN [Reduced_T_Rappro_N - 1] ON (
    Reduced_T_Rappro_N.[Rente RES] = [Reduced_T_Rappro_N - 1].[Rente RES]
  ) 
  AND (
    Reduced_T_Rappro_N.[Société] = [Reduced_T_Rappro_N - 1].[Société]
  ) 
GROUP BY 
  Reduced_T_Rappro_N.[Rente RES], 
  Reduced_T_Rappro_N.[Montant capital constitutif], 
  [Reduced_T_Rappro_N - 1].[Montant capital constitutif], 
  [Reduced_T_Rappro_N].[Montant Sous Rente], 
  [Reduced_T_Rappro_N - 1].[Montant Sous Rente], 
  Reduced_T_Rappro_N.[Montant rente initial], 
  [Reduced_T_Rappro_N - 1].[Montant rente initial];

我查询中的划分对应于百分比进化的计算，此外我使用 MS ACCESS 从那里调用方法 iif

结果：

Rente_RES	Montant_Capital_Consitutif_N	Montant_Capital_Consitutif_N-1	Montant Sous Rente	Montant Sous Rente N-1	Evolution montant sous rente	Montant Rente Initial N	Montant Rente initial N-1	Addition
00000002-01	0	0	200,34	198,35	1,00327703554324	195,61	195,61	398,69
00000002-01	0	0	200,34	200,34	0	195,61	195,61	400,68
00000002-01	0	0	202,34	198,35	2,0115956642299	195,61	195,61	400,68
00000002-01	0	0	202,34	200,34	0,998302885095338	195,61	195,61	402,68
00000002-03	25070,68	25070,68	1583,2	1583,2	0	1472,18	1472,18	3166,4
00000003-04	0	0	1358,5	1340,93	1,31028465318845	1266,7	1266,7	2699,43
00000003-04	0	0	1358,5	1358,5	0	1266,7	1266,7	2717

我想要的： [Montant sous rente N] 和 [Montant sous rente N-1] 之间的加法最大值对应的每个 ID (Rente RES) 只获得一个结果。

Rente_RES	Montant_Capital_Consitutif_N	Montant_Capital_Consitutif_N-1	Montant Sous Rente	Montant Sous Rente N-1	Evolution montant sous rente	Montant Rente Initial N	Montant Rente initial N-1	Addition
00000002-01	0	0	202,34	200,34	0,998302885095338	195,61	195,61	402,68
00000002-03	25070,68	25070,68	1583,2	1583,2	0	1472,18	1472,18	3166,4
00000003-04	0	0	1358,5	1358,5	0	1266,7	1266,7	2717

Answer 1

SELECT Reduced_T_Rappro_N.Rente_RES, Max(([Montant_Capital_Consitutif_N-1]+[Montant_Capital_Consitutif_N])) 作为加法来自 Reduced_T_Rappro_N 分组 Reduced_T_Rappro_N.Rente_RES;

获得一对 table 的所有 ID 的最大值

get the max value of a couple for all ids of a table

sql

ms-access