Question

我有两个桌子

表A

+-----+---------+---------+
| pid | name    | country |
+-----+---------+---------+
|  1  | ABC     | XXXXXXX |
|  2  | EFG     | YYYYYYY |
|  3  | IJK     | ZZZZZZZ |
|  4  | LMN     | AAAAAAA |
|  5  | OPQ     | BBBBBBB |
|  6  | RST     | CCCCCCC |
+-----+---------+---------+

表B

+-----+---------+
| id  | pid     |
+-----+---------+
|  5  |   5     | 
|  4  |   1     | 
|  3  |   2     | 
|  2  |   5     | 
|  1  |   2     |
+-----+---------+

我想要下面的输出

+-----+---------+---------+
| pid | name    | country |
+-----+---------+---------+
|  5  | OPQ     | BBBBBBB |
|  1  | ABC     | XXXXXXX |
|  2  | EFG     | YYYYYYY |
+-----+---------+---------+

这是输出应获取表B id字段的最新条目，并且表A的数据应显示在表B的pid出现顺序上

Answer 1

尝试一下：

SELECT DISTINCT A.PID, A.NAME, A.COUNTRY
FROM TABLE_A A
INNER JOIN TABLE_B B
ON A.PID = B.PID
ORDER BY A.PID

Answer 2

您需要将表连接在一起并获取最新值。这是一种方法：

select a.pid, a.name, a.country
from a join
     b
     on a.pid = b.pid
where b.id = (select max(b2.id) from b b2 where b2.pid = b.pid);

Answer 3

您不需要B的联接或字段吗？

SELECT a.pi, a.name, a.country
FROM A AS a
WHERE EXISTS(SELECT b.id FROM B AS b WHERE b.pid = a.pid)

Answer 4

您也可以使用class currencyAccounts: Object { @objc dynamic var currencyID : Int = 0 @objc dynamic var currencyName = "" @objc dynamic var currencyCode = "" @objc dynamic var currencyBalance : Double = 0.00 @objc dynamic var currencyStatusCode : Bool = false override static func primaryKey() -> String? { return "currencyID" } convenience init(currencyID: Int, currencyName: String, currencyCode: String, currencyStatusCode: Bool, currencyBalance: Double) { self.init() self.currencyID = currencyID self.currencyName = currencyName self.currencyCode = currencyCode self.currencyStatusCode = currencyStatusCode self.currencyBalance = currencyBalance }：

row_number()

选择按ID排序的唯一行

4 个答案: