Question

我有三张表如下：

第一张表：

ordenes

func downloadFaceId() {
    if let image = personimage, let imageData = UIImageJPEGRepresentation(image, 0.8) {
        FaceService.instance.client.detect(with: imageData, returnFaceId: true, returnFaceLandmarks: false, returnFaceAttributes: nil, completionBlock: { (face:[MPOFace]!, err:NSError!) in
            if err == nil {
                var faceId: String?
                for face in faces {
                    faceId = face.faceId
                    break
                }

                self.faceId = faceId
            }
         })

第二张表：

usuario

id_orden | date | total | id_usuario
1        |15-may|50     | 1
2        |20-may|60     | 2

第三张表：

埃斯塔

id_usuario | name | phone
1          | abc  | 999
2          | def  | 888

这是理想的结果：

结果：

id_orden | edo
1        | c
1        | b
1        | a
2        | b
2        | a

maxedo需要是基于订单汇总后第三个表中edo的最大记录。

我该怎么做？

Answer 1

以下代码示例为您提供结果。

CREATE TABLE #ordenes(id_orden int, datevalue date, total int, id_usuario int)
INSERT INTO #ordenes
VALUES
(1,'20160515',50,1),
(2,'20160520',60,2)

CREATE TABLE #usuario(id_usuario int, name varchar(10), phone int)
INSERT INTO #usuario
VALUES 
(1,'abc',999),
(2,'def',888)

CREATE TABLE #estado(id_orden int, edo char(1))
INSERT INTO #estado
VALUES
(1,'c'),
(1,'b'),
(1,'a'),
(2,'b'),
(2,'a')

SELECT id_orden,datevalue,total,id_usuario,name,phone,edo as maxedo
FROM 
(SELECT o.id_orden,o.datevalue,o.total,o.id_usuario,u.name,u.phone,e.edo,ROW_NUMBER() OVER(PARTITION BY o.id_orden ORDER BY e.edo DESC) as rnk
FROM #ordenes o
JOIN #usuario u
on o.id_usuario = u.id_usuario
join #estado e
on o.id_orden = e.id_orden) as t
where rnk = 1

Answer 2

以下应该做的工作（假设edo实际上是一个数字量）。我使用AS命令包含了别名，因此您甚至可以获得所需的列标题。

SELECT
    oe.id_orden AS id_orden,
    oe.date AS date,
    oe.total AS total,
    u.id_usario AS id_usuario,
    u.name AS name,
    u.phone AS phone,
    oe.maxedo AS maxedo
FROM usuario u
INNER JOIN 
(SELECT 
    o.id_orden AS id_orden,
    o.date AS date,
    o.total AS total,
    o.id_usuario AS id_usuario,
    e.maxestedo AS maxestedo
FROM ordenes o
INNER JOIN
(SELECT 
    id_orden AS id_orden,
    MAX(edo) AS maxedo
FROM estado
GROUP BY id_orden) e
ON e.id_orden=o.id_orden) oe
ON u.id_usuario=oe.id_usuario

为了处理（这不是SQL如何工作，但是将其分解为步骤的有用方法），它会：

创建最大edos表（NB：MAX也适用于字母顺序）;
使用id_ordene;

ordenes

使用usuario将此加入id_usuario数据;和
以所需格式将其发布为表格。

Answer 3

问题可以分为以下三个步骤：

第1步：计算表estado中每个id_orden的最大edo：

Select id_orden, max(edo) maxedo
From estado
Group By id_orden;

结果：

| id_orden | edo |
|   1      |  c  |
|   2      |  b  |

第2步：在关键字“id_usuario”上加入两个表ordenes和usuario：

Select o.id_orden, o.date, o.total, o.id_usuario, u.name, u.phone 
From ordenes o Join usuario u
On o.id_usuario = u.id_usuario;

结果：

id_orden | date | total | id_usuario | name | phone 
1        |15-may|50     | 1          | abc  | 999   
2        |20-may|60     | 2          | def  | 888

第3步：在关键字id_orden上加入表格step1和step2：

Select a.id_orden, a.date, a.total, a.id_usuario, a.name, a.phone, b.maxestado
From (Select o.id_orden, o.date, o.total, o.id_usuario, u.name, u.phone 
From ordenes o Inner Join usuario u
On o.id_usuario = u.id_usuario ) a 
Join (Select id_orden, max(edo) maxestado
From estado
Group By id_orden) b
On a.id_orden = b.id_orden;

结果：

id_orden | date | total | id_usuario | name | phone | maxedo
1        |15-may|50     | 1          | abc  | 999   |    c
2        |20-may|60     | 2          | def  | 888   |    b

SQLFiddle示例：http://sqlfiddle.com/#!5/a79a1/2

：）

Answer 4

我认为您需要通过id_orden从第三个表和组中获取max（edo），是吗？试试这个。

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如何内连三个表并返回第三个最大值

4 个答案: