Question

我已经创建了手机游戏，我将得分数据（在这种情况下为矿物）存储在sql表中。现在，我想通过从两天前的分数中减去昨天的分数来创建每日分数图表。我发现我可以用EXCEPT做到这一点，但是我没有尝试。

    $sql = "SELECT playerid, playername, minerals, daydate
    FROM dailyscore WHERE daydate = '".$yesterday."'
    EXCEPT
    SELECT playerid, playername, minerals, daydate
    FROM dailyscore WHERE daydate = '".$twodays."'
    ORDER BY minerals DESC";

我收到此错误：SQL错误：“您的SQL语法有错误；请查看与您的MySQL服务器版本相对应的手册，以获取正确的语法，以在'EXCEPT SELECT玩家编号，玩家名称，矿物，daydate FROM Dailyscore附近使用da'在第3行的位置”

Answer 1

不需要像EXCEPT或MINUS这样的集合运算符。只需将两天的分数相加即可计算出每日分数的差异：

SELECT playerid, playername, minerals
FROM (
    SELECT ds_yesterday.playerid,
        ds_yesterday.playername,
        CASE
            WHEN ds_twodays.minerals IS NULL
            THEN ds_yesterday.minerals
            ELSE ds_yesterday.minerals - ds_twodays.minerals
        END minerals
    FROM dailyscore as ds_yesterday
    LEFT JOIN dailyscore as ds_twodays ON 
        ds_twodays.playerid = ds_yesterday.playerid
        AND ds_twodays.daydate = '2018-08-07'
    WHERE ds_yesterday.daydate = '2018-08-08'
) scoredifference
ORDER BY minerals DESC;

您可以使用此SQL并将其放在您的PHP代码中。日期应替换为存储在PHP变量中的值。

Answer 2

MySQL不支持EXCEPT。我认为您有用：

SELECT ds.playerid, ds.playername, ds.minerals, ds.daydate
FROM dailyscore ds
WHERE daydate = '".$yesterday."' AND
      NOT EXISTS (SELECT 1
                  FROM dailyscore
                  WHERE ds2.playerid = ds.playerid AND
                        ds2.minerals = ds.minerals AND
                        ds2.daydate = '".$twodays."'
                 )
ORDER BY ds.minerals DESC;

注意：您不应该通过修饰字符串来传递参数。您应该学习使用?之类的占位符正确地传递参数。

如何从一个SQL表中的数据减去PHP中的另一个SQL表？

2 个答案: