通过内部联接选择

Question

我需要从两个表中选择一些数据，请帮助我使用内部联接进行此选择。 选择2中的玩家一定不能进入选择1 ...首先选择：

$rs = "SELECT * 
        FROM `player` 
        WHERE `status`=1 AND `credit`>=1 AND `username` NOT LIKE '$user' 
        ORDER BY ls ASC,credit DESC 
        LIMIT 0 ,10;

第二：该玩家必须从选择结果中删除1

$rs2 = "SELECT * 
        FROM `ip_log` 
        WHERE `playerid`='$ui' AND `win`='1' AND `date`='$date' ";`

Answer 1

您可以为此使用LEFT JOIN ：

这将显示未选择1的每个人的日志消息。

SELECT l.*
FROM ip_log AS l
LEFT JOIN 
    (SELECT username
     FROM player
     WHERE status = 1 AND credit >= 1 AND username NOT LIKE '$user'
     ORDER BY ls ASC, credit DESC 
     LIMIT 10) AS p
ON l.player = p.username
WHERE win = 1 and date = '$date'
AND p.username IS NULL

这显示了前10个玩家数据，选择2中带有日志消息的数据除外

SELECT p.*
FROM player AS p
LEFT JOIN ip_log AS l ON l.player = p.username AND l.win = 1 AND l.date = '$date'
WHERE p.status = 1 AND p.credit >= 1 AND p.username NOT LIKE '$user'
AND l.player IS NULL
ORDER BY p.ls ASC, p.credit DESC
LIMIT 10

在这两种情况下，使用IS NULL测试第二个表中的列都会使其仅返回第一个表中与第二个表中不匹配的行。 看到

仅当值不存在时才返回行

Answer 2

您可以使用LEFT JOIN完成

SELECT player.*,ip_log.* FROM `player` LEFT JOIN `ip_log` ON player.id!=ip_log.playerid GROUP BY player.id