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我会尽力解释这一点。我有两个从 MySQL 中提取数据的查询，一个是关于公园评论的数据，另一个是关于乘车评论的数据。两个查询都按 排序Review Date。

我在网页上想要的是一份按排序的评论列表，Review Date因此公园评论和骑行评论将在一个列表中混合在一起。

几乎就像在运行两个 foreach 循环，每个查询一个，但根据日期轮流执行。

这是两个查询：对于游乐设施

$query4= 'SELECT review_id, review, tpf_reviews_rides.user, DATE_FORMAT(date_ridden, "%d %M %Y") AS Date_Ridden, DATE_FORMAT(review_date, "%d %M %Y") AS Review_Date, tpf_rides.name AS ride_name, rating, tpf_parks.name  AS park_name, country
FROM tpf_reviews_rides  

INNER JOIN tpf_rides ON tpf_reviews_rides.ride_id = tpf_rides.ride_id
INNER JOIN tpf_ratings_rides ON tpf_reviews_rides.rating_link = tpf_ratings_rides.rating_id  
INNER JOIN tpf_parks ON tpf_reviews_rides.park_id = tpf_parks.park_id

ORDER BY review_date DESC, review_id DESC LIMIT '.$start_from.' , '.$limit.'';

对于公园：

$query5= 'SELECT review_id, review, tpf_reviews_parks.user, DATE_FORMAT(date_visited, "%d %M %Y") AS Date_Visited, DATE_FORMAT(review_date, "%d %M %Y") AS Review_Date, rating, tpf_parks.name, country
FROM tpf_reviews_parks  

INNER JOIN tpf_ratings_parks ON tpf_reviews_parks.rating_link = tpf_ratings_parks.rating_id  
INNER JOIN tpf_parks ON tpf_reviews_parks.park_id = tpf_parks.park_id

ORDER BY review_date DESC, review_id DESC LIMIT '.$start_from.' , '.$limit.'';

这就是我分别显示结果的方式：对于游乐设施：

foreach ($res4 AS $row4) { 
            printf('
            <h2 style="display:inline;">%s</h2> <h3 style="display:inline;">- %s, %s</h3>
            <h3>Rating - %sStars</h3>
            <h4>Submitted by %s on %s</h4>
            <p>%s</p>' . PHP_EOL, $row4['ride_name'], $row4['park_name'], $row4['country'], $row4['rating'], (htmlspecialchars($row4['user'], ENT_QUOTES, 'UTF-8')), $row4['Review_Date'], nl2br(htmlspecialchars($row4['review'], ENT_QUOTES, 'UTF-8')));

            if(!empty($row4['Date_Ridden'])){
            printf('<p>Date ridden: %s</p>' . PHP_EOL, $row4['Date_Ridden']);
            }

            printf('<a class="bloglink" href="#top">Back to Top</a>
            <hr>');

        }


    For PArks:

foreach ($res5 AS $row5) { 
    printf('
    <h2 style="display:inline;">%s</h2> <h3 style="display:inline;">- %s</h3>
    <h3>Rating - %sStars</h3>
    <h4>Submitted by %s on %s</h4>
    <p>%s</p>' . PHP_EOL, $row5['name'], $row5['country'], $row5['rating'], (htmlspecialchars($row5['user'], ENT_QUOTES, 'UTF-8')), $row5['Review_Date'], nl2br(htmlspecialchars($row5['review'], ENT_QUOTES, 'UTF-8')));

    if(!empty($row4['Date_Ridden'])){
    printf('<p>Date Visited: %s</p>' . PHP_EOL, $row4['Date_Visited']);
    }

    printf('<a class="bloglink" href="#top">Back to Top</a>
    <hr>');

}

为了使它更棘手，这个页面是分页的。看看它，因为这两个查询非常相似，如果有一种方法可以在查询级别组合，那将是最好的，并有助于保持分页代码的完整。有人对我如何实现这一目标有任何建议吗？

谢谢

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1 回答 1

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您可以使用UNION将两个 select 语句合并在一起。

请注意，如果您的表非常相似，最好将它们合并（不是在查询中，而是在实际表中）。tpf_reviews_rides我没有看到任何与您相比不完全相同的字段tpf_reviews_parks。添加一个表示评论类型（乘车或停车）的列，您应该进行设置。它将简化您的查询和处理逻辑。

SELECT
    type,
    review_id,
    review,
    user,
    Date_Event,
    Review_Date,
    ride_name,
    rating,
    park_name,
    country
FROM (
    SELECT
        'ride' as type,
        review_id,
        review,
        tpf_reviews_rides.user,
        DATE_FORMAT(date_ridden, "%d %M %Y") AS Date_Event,
        DATE_FORMAT(review_date, "%d %M %Y") AS Review_Date,
        tpf_rides.name AS ride_name,
        rating,
        tpf_parks.name AS park_name,
        country
    FROM
        tpf_reviews_rides  
        INNER JOIN tpf_rides ON tpf_reviews_rides.ride_id = tpf_rides.ride_id
        INNER JOIN tpf_ratings_rides ON tpf_reviews_rides.rating_link = tpf_ratings_rides.rating_id  
        INNER JOIN tpf_parks ON tpf_reviews_rides.park_id = tpf_parks.park_id

    UNION

    SELECT
        'park' as type,
        review_id,
        review,
        tpf_reviews_parks.user,
        DATE_FORMAT(date_visited, "%d %M %Y") AS Date_Event,
        DATE_FORMAT(review_date, "%d %M %Y") AS Review_Date,
        NULL as ride_name,
        rating,
        tpf_parks.name AS park_name,
        country
    FROM
        tpf_reviews_parks  
        INNER JOIN tpf_ratings_parks ON tpf_reviews_parks.rating_link = tpf_ratings_parks.rating_id  
        INNER JOIN tpf_parks ON tpf_reviews_parks.park_id = tpf_parks.park_id
) AS reviews
ORDER BY
    review_date DESC,
    review_id DESC
LIMIT ' . $start_from . ', ' . $limit;
于 2013-08-19T19:36:11.253 回答