Question

我有两张桌子：

第一个包含当天（每小时）的数据 - 此表包含需要处理的原始数据。
第二个包含前一天（每天）的数据 - 此表已包含每天计算的值。

我想将它与MySQL查询中的行组合在一起，以便我返回之前日期（每天）和当天（再次，每天）的数据。

目前我正在使用2个mysql查询：

//table 1, data for current day
                    $qry1="
            select DATE_FORMAT(completedate,'%Y-%m-%d') as date, SUM(IF(complete = 1,affpayout,0)) as pay, COUNT(*) as leads
            from leads
            where affiliateid={$_SESSION["id"]}  AND completedate>'$date' 
            GROUP BY DATE_FORMAT(completedate,'%Y-%m-%d')";
//table 2, data for previous days, already processsed, we just need to select it
            $qry2="
            select DATE(date) as date, affrevenue as pay, totalleads as leads
            from leadsdays
            GROUP BY DATE(date)";

如何有效地结合2（速度性能是一个问题）？最好的方法是什么？

Answer 1

尝试UNION。此外，请确保您的表已正确索引。看起来您至少需要completedate和affiliateid索引。不确定两个查询与一个联盟相比效率更高......

$qry = "(select 
    DATE_FORMAT(completedate,'%Y-%m-%d') as `date`, 
    SUM(IF(complete = 1,affpayout,0)) as pay, 
    COUNT(*) as leads
    from leads
    where affiliateid={$_SESSION["id"]}  AND completedate>'$date' 
    GROUP BY DATE_FORMAT(completedate,'%Y-%m-%d')) 
    UNION 
    (select 
    DATE(`date`) as `date`, 
    affrevenue as pay, 
    totalleads as leads
    from leadsdays
    GROUP BY DATE(`date`)) 
    ORDER BY DATE(`date`)";

将2个Mysql查询合并为1（高效方式）

1 个答案: