首页 > 解决方案 > 在 MS Access 中对每个查询结果运行子查询

问题描述

我目前在 PHP 应用程序中使用此 SQL,需要转换为 MS Access。

我需要针对第一个结果中的每个结果运行第二个查询,以最终计算出第一场比赛中的骑手参加的每场比赛中的骑手人数。(摩托车比赛项目)

使用 PHP 非常简单(如果需要,我可以提供完整的代码)

甚至不确定如何从 Access 开始。(两个查询在 Access 中都可以正常工作,只是无法按照我的需要加入它们)

查询一:

SELECT schedule2.Date, riders.ridername, results.finish, results.points, classes.class, seasons.season
FROM schedule2 RIGHT JOIN (classes RIGHT JOIN ((riders RIGHT JOIN results ON riders.id = results.rider_id) LEFT JOIN seasons ON results.season_id = seasons.id) ON classes.id = results.class_id) ON schedule2.id = results.schedule_id
GROUP BY schedule2.Date, riders.ridername, results.finish, results.points, classes.class, seasons.season, schedule2.id, results.class_id, riders.id
HAVING (((riders.id)=[:id]))
ORDER BY schedule2.Date;

查询 2:

SELECT DISTINCT Count(results.rider_id) AS CountOfrider_id
FROM seasons RIGHT JOIN (results LEFT JOIN schedule2 ON results.schedule_id = schedule2.id) ON seasons.id = schedule2.season_id
GROUP BY schedule2.id, results.class_id
HAVING (((schedule2.id)=[:scheduleId]) AND ((results.class_id)=[:classId]));

当前的 PHP 应用程序生成这个结果集:(使用表单选择骑手 id)

在此处输入图像描述

原始PHP(供参考)

            $getRiderInfo = $dbh->prepare('
                    SELECT riders.ridername, 
                          schedule2.date, 
                          schedule2.id AS scheduleId, 
                          results.class_id AS classId, 
                          results.finish, 
                          results.points, 
                          classes.class, 
                          seasons.season, 
                          riders.id 
                    FROM schedule2 
                    RIGHT JOIN (classes 
                                RIGHT JOIN ((riders 
                                             RIGHT JOIN results 
                                             ON riders.id = results.rider_id) 
                    LEFT JOIN seasons 
                      ON results.season_id = seasons.id) 
                      ON classes.id = results.class_id) 
                      ON schedule2.id = results.schedule_id
                    GROUP BY riders.ridername, 
                             schedule2.date, 
                             classes.class, 
                             seasons.season
                    HAVING (((riders.id)=:id))
                    ORDER BY schedule2.date ASC;
            ');
            $getRiderInfo->bindValue('id', $racer_id);
            if ($getRiderInfo->execute()) {
                while ($iRows = $getRiderInfo->fetch(PDO::FETCH_ASSOC)) {
                    $getCounts = $dbh->prepare('
                            SELECT DISTINCT COUNT(results.rider_id) AS CountOfrider_id,
                                   schedule2.id,
                                   results.class_id
                            FROM seasons
                            RIGHT JOIN(results
                            LEFT JOIN schedule2
                            ON results.schedule_id = schedule2.id) 
                            ON seasons.id = schedule2.season_id
                            GROUP BY schedule2.id , 
                                     results.class_id
                            HAVING (((schedule2.id) = :scheduleId)
                            AND ((results.class_id) = :classId));
                    ');
                    $getCounts->bindValue('scheduleId', $iRows['scheduleId']);
                    $getCounts->bindValue('classId', $iRows['classId']);
                    if ($getCounts->execute()) {
                        while ($iRows1 = $getCounts->fetch(PDO::FETCH_ASSOC)) {
                            $iCount = $iRows1['CountOfrider_id'];
                        }
                    }
                    $iDate = date_create($iRows['date']);
                    echo '<tr><td>' . $iRows['ridername'] . '</td>';
                    echo '<td>' . date_format($iDate, 'm/d/Y') . '</td>';
                    echo '<td>' . $iRows['finish'] . '</td>';
                    echo '<td>' . $iRows['points'] . '</td>';
                    echo '<td>' . $iRows['class'] . '</td>';
                    echo '<td>' . $iRows['season'] . '</td>';
                    echo '<td>' . $iCount . '</td></tr>';
                }
            }

标签: sqlms-access-2016

解决方案


我找到了解决这个问题的更好方法。其实我有点想多了。

这个数据集永远不会改变,因此,第二个查询并不是真正需要的,只需运行一次以创建一个表,其中包含我正在寻找的第 6 列的总数,并在第一个查询中添加了另一个连接。(已经有太多的连接无法跟踪。)

它适用于我的目的。


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