MySQL之間存在的第一個空閒數值

Question

問題：在存在值之間爲服務器獲取第一個空閒端口。如果沒有空閒的地方，則取最高+ 1.

額外：如果server_deleted = 1，那麼我們可以採用空閒端口。當價值之間存在差距時，我們也應該採取自由港。我們應該在where子句中包含server_type。最小值是第一個端口，例如現在9000

我的查詢如下：

SELECT server_port + 1 
    FROM pro_servers s 
WHERE s.server_port <> 0 
    AND s.server_type = 'ts3' 
    AND s.server_deleted = 0 
    AND NOT EXISTS 
     (SELECT s1.server_port 
      FROM pro_servers s1 
      WHERE s1.server_port <> 0 
      AND s1.server_type = 'ts3' 
      AND s1.server_port = s.server_port + 1 
      AND s1.server_deleted = 0 
     ) 
ORDER BY server_port LIMIT 1 

我認爲有更好的方式來做到這一點。此查詢執行速度非常慢。

例如first = minimum = lowest = 9000，接下來是9002,9003.我們需要獲得9001.如果我們添加9001，那麼得到9004.第一個值9000存在於表中。

樣本數據

+-----------+-------------+----------------+-------------+ 
| server_id | server_port | server_deleted | server_type | 
+-----------+-------------+----------------+-------------+ 
|  151 |  9500 |    1 | teamspeak3 | 
|  8459 |  9500 |    0 | teamspeak3 | 
|  183 |  9501 |    1 | teamspeak3 | 
|  264 |  9502 |    1 | teamspeak3 | 
|  4155 |  9502 |    1 | teamspeak3 | 
|  2707 |  9503 |    1 | teamspeak3 | 
|  4160 |  9503 |    1 | teamspeak3 | 
|  154 |  9504 |    1 | teamspeak3 | 
|  4163 |  9504 |    1 | teamspeak3 | 
|  285 |  9506 |    1 | teamspeak3 | 
|  4167 |  9506 |    1 | teamspeak3 | 
|  8454 |  9506 |    0 | teamspeak3 | 
|  241 |  9507 |    1 | teamspeak3 | 
|  4169 |  9507 |    1 | teamspeak3 | 
|  188 |  9509 |    1 | teamspeak3 | 
|  4177 |  9509 |    1 | teamspeak3 | 
+-----------+-------------+----------------+-------------+ 

查詢結果：9501當我們使用這個端口，然後下一個：9502，9503，9504，9505，9507，9508，9509，9510等

Answer 1

SELECT data.sPort 
FROM 
    ((SELECT (s.server_port + 1) sPort 
    FROM pro_servers s 
    LEFT JOIN pro_servers sp1 ON sp1.server_port = s.server_port + 1 
    WHERE (sp1.server_port IS NULL) 
    ORDER BY sPort) 

    UNION ALL 

    (SELECT s.server_port sPost 
    FROM pro_servers s 
    GROUP BY s.server_port 
    HAVING COUNT(s.server_port) = SUM(s.server_deleted) 
    ORDER BY sPort)) AS data 
ORDER BY data.sPort 
LIMIT 1 

SqlFiddle：http://sqlfiddle.com/#!2/12ab1/2

它是如何工作

1. 左連接pro_servers與pro_servers與server_port = server_port + 1連接條件，並採取行，其中有在下一個端口null。這些行顯示每個間隙的第一個端口。端口ID可以被視爲server_port + 1。

2. 取出所有已刪除的端口。

3. 聯合1.和2.，訂購併採取第一個。

對於答案有一個假設 - 總是採用編號最小的端口。如果不是這樣，請單獨檢查該端口（或將另一個UNION ALL添加到查詢中）。

Answer 2

此外，您可以嘗試索引表。

CREATE INDEX indexName ON tableName（field1 [，field2 ...]）;

Answer 3

我會用這個查詢使用NOT EXISTS：

SELECT MIN(server_port)+1 
FROM pro_servers p1 
WHERE 
    p1.server_type = 'ts3' AND 
    NOT EXISTS (SELECT server_port 
       FROM pro_servers p2 
       WHERE p1.server_port=p2.server_port-1 
        AND p1.server_type=p2.server_type 
        AND p2.server_deleted=0) 

或這是使用LEFT JOIN：

SELECT MIN(p1.server_port)+1 
FROM 
    pro_servers p1 LEFT JOIN pro_servers p2 
    ON p1.server_port=p2.server_port-1 
    AND p1.server_type=p2.server_type 
    AND p2.server_deleted=0 
WHERE 
    p2.server_port IS NULL 
    AND p1.server_type='ts3' 

請參閱小提琴here。

Answer 4

我不知道這是否會更快，但它肯定會幫助你避免再犯同樣的條件下兩次：

SELECT 
    MAX(server_port) + 1 AS first_available_port 
FROM (
    SELECT 
    server_port, 
    @row := @row + 1 AS row 
    FROM 
    pro_servers AS s, 
    (SELECT @row := 0) AS x 
    WHERE server_port <> 0 
    AND server_type = 'ts3' 
    AND server_deleted = 0 
) AS s 
GROUP BY 
    server_port - row 
ORDER BY 
    server_port - row 
LIMIT 1 
; 

內部查詢列舉符合條件的現有端口。現在，對於屬於同一組連續行的行，端口號和行號之間的差異將保持不變。外部查詢按該差異進行分組，並返回第一組的最高端口加1。