Php 如何在html表中填充UNION ALL查询数据

如何使用PHP和MYSQL获取的两侧标题填充html表中的数据。我没有得到想要的输出显示，就像在所需的输出图像中一样

$query = $db->prepare("SELECT player, score FROM cricket_table WHERE dateby BETWEEN '2020-01-01' AND '2020-03-01'
UNION ALL
SELECT player, score FROM football_table WHERE dateby BETWEEN '2020-01-01' AND '2020-03-01'");
$query->execute; 
echo '<table><tr><th>Player Name</th><th>Cricket</th><th>Football</th></tr>';
while($row = $query->fetch(PDO:: FETCH_ASSOC) ) {
echo '<tr>
        <td>'.$row['player'].'</td>
        <td>'.$row['score'].'</td>
     </tr>';
}
echo '</table>';

这是我们想要的

这是电流输出

现在尝试输出表格

while($row = $query->fetch(PDO:: FETCH_ASSOC) ) {
echo '<tr>
        <td>'.$row['player'].'</td>
        <td>'.$row['cr_score'].'</td>
        <td>'.$row['fb_score'].'</td>
     </tr>';
}

---

无法对其进行测试，但您得到了大致思路，应该可以使用UNION ALL并对每列所需的值求和：

这是您的SQL。注意，它将使用union all运行两个查询，然后对每个值求和。我们将使用0零技巧忽略错误的表：

SELECT x.player, SUM(x.cricket_score) as cricket_score, SUM(x.football_score) as football_score FROM (
    SELECT player, SUM(score) as cricket_score, 0 as football_score
      FROM cricket_table
     WHERE dateby BETWEEN '2020-01-01' AND '2020-03-01'
     GROUP BY player
     UNION ALL
    SELECT player, 0 as cricket_score, SUM(score) as football_score
      FROM football_table 
     WHERE dateby BETWEEN '2020-01-01' AND '2020-03-01'
     GROUP BY player
) x
GROUP BY x.player

并适用于该守则：

$query = $db->prepare("    SELECT x.player, SUM(x.cricket_score) as cricket_score, SUM(x.football_score) as football_score FROM (
    SELECT player, SUM(score) as cricket_score, 0 as football_score
      FROM cricket_table
     WHERE dateby BETWEEN '2020-01-01' AND '2020-03-01'
     GROUP BY player
     UNION ALL
    SELECT player, 0 as cricket_score, SUM(score) as football_score
      FROM football_table 
     WHERE dateby BETWEEN '2020-01-01' AND '2020-03-01'
     GROUP BY player
) x
GROUP BY x.player");
$query->execute();
echo '<table><tr><th>Player Name</th><th>Cricket</th><th>Football</th></tr>';
while($row = $query->fetch(PDO:: FETCH_ASSOC) ) {
    echo '<tr>
        <td>'.$row['player'].'</td>
        <td>'.$row['cricket_score'].'</td>
        <td>'.$row['football_score'].'</td>
     </tr>';
}
echo '</table>';

---

使用此SQL，您将获得cricket或football已经设置的值。

除了您提供的代码无效之外，您还有什么问题？我没有得到我想要的输出显示，如上面所需的输出图像。如果不同表中有相同的列，为什么不只使用一个表和标志？例如，在WordPress中，他们有wp_posts表和post_type，其中存储了post的类型，例如：产品、附件、post等。您可以在旗帜中的位置执行相同的操作，例如：sport_type您可以存储值，如cricket，足球等。然后你可以使用左JOINUNION，因为返回的结果就像你只查询一个表一样。您必须使用外键连接，外键可以是player\u id：从players中选择a.*，c.*，f.*作为左连接cricket\u表作为a.player\u id=c.player\u id左连接football\u表作为f作为a.player\u id=f.player\u id请提供真实的示例数据，或者如果运动员不一定有每项运动的分数，您可以使用case和nulls，或者别的什么，但这是一个简单的版本，我试过这个方法。它返回了错误的结果，并且提交了较大的日期范围，如2年之间。mysql服务器停止工作。通过结束mysql进程，并给出以下错误警告：未捕获PDO异常：PDO:：_构造：mysql服务器已在union中消失，它返回正确的结果，但未正确填充到html表中。它返回板球和足球的板球数据。比如如果板球得分是44，那么足球的得分也是44。每一列的数据来源是不同的。您确定使用的是完全相同的代码，和/或表具有不同的数据吗？我不明白它是如何发生的，正如你所描述的。是的，我使用的是精确的方法，所有的表都有彼此不同的数据。它完全按照预期工作：。。您确定使用的是相同的代码吗？如果有更多的表，则需要为列指定不同的名称，并在所有列中使用0零技巧。这取决于数据的存储方式。您可以尝试添加时间部分，以确保它将获得日内小时数，如2020-05-03 00:00:00和2020-05-03 23:59:59

$query = $db->prepare("    SELECT x.player, SUM(x.cricket_score) as cricket_score, SUM(x.football_score) as football_score FROM (
    SELECT player, SUM(score) as cricket_score, 0 as football_score
      FROM cricket_table
     WHERE dateby BETWEEN '2020-01-01' AND '2020-03-01'
     GROUP BY player
     UNION ALL
    SELECT player, 0 as cricket_score, SUM(score) as football_score
      FROM football_table 
     WHERE dateby BETWEEN '2020-01-01' AND '2020-03-01'
     GROUP BY player
) x
GROUP BY x.player");
$query->execute();
echo '<table><tr><th>Player Name</th><th>Cricket</th><th>Football</th></tr>';
while($row = $query->fetch(PDO:: FETCH_ASSOC) ) {
    echo '<tr>
        <td>'.$row['player'].'</td>
        <td>'.$row['cricket_score'].'</td>
        <td>'.$row['football_score'].'</td>
     </tr>';
}
echo '</table>';