Question

我正在研究从学校毕业的项目，但现在我遇到了一个问题。我的两个问题是：

$sql = 'SELECT Amount FROM activities ORDER BY Id ASC';

和

$select = 'SELECT COUNT(Activity) as Activity 
           FROM `registrations` 
           INNER JOIN activities 
                ON registrations.Activity = activities.Name 
           WHERE registrations.Activity='.$row['Name'].' 
                AND registrations.Morning/Afternoon='.$row['Morning/Afternoon'].'';

所以需要发生的是$ sql查询中的金额 - 来自$ select查询的COUNT（活动）。

到目前为止，我得到了这个，但它并不像我希望的那样工作：

<?php

$sql = 'SELECT * FROM activities ORDER BY Id ASC';
foreach ($db->query($sql) as $row) {
    $select = 'SELECT COUNT(Activity) as Activity 
               FROM `registrations` 
               INNER JOIN activities 
                  ON registrations.Activity = activities.Name 
               WHERE registrations.Activity=' . $row['Name'] . ' 
                  AND registrations.Morning/Afternoon=' . $row['Morning/Afternoon'] . '';

    foreach ((array)$select as $select2) {
        $maxAantal = $row['Amount'] - $select2['Activity'];
    }

    echo '<tr>';
    echo '<td>' . $row['Name'] . '</td>';
    echo '<td>' . $row['Location'] . '</td>';
    echo '<td>' . $row['Begintime'] . '</td>';
    echo '<td>' . $row['Endtime'] . '</td>';
    echo '<td>' . $maxAmount . '</td>';
    echo '<td>' . $row['Deadline'] . '</td>';
    echo '<td>' . $row['Morning/Afternoon'] . '</td>';
    echo '<td>';
    echo '<a class="btn btn-secondary" href="readActi.php?id=' . $row['Id'] . '">Read</a>';
    echo '</td>';
    echo '</tr>';
}

$_SESSION['Id'] = $row['Id'];

?>

但它给了我这个警告

Warning: Illegal string offset 'Activity'

我无法弄清楚如何解决这个问题

Answer 1

您的第二个查询$select未执行也未获取，只会在变量$select中设置...您只是忘记查询并获取它，这就是为什么$select2['Activity']未设置的原因

检查您的第二个foreach，我认为它应该是foreach ($db->query($select) as $select2)

精确编辑

在您的代码中，您有：

$select = 'SELECT COUNT(Activity) as Activity 
           FROM `registrations` 
           INNER JOIN activities 
              ON registrations.Activity = activities.Name 
           WHERE registrations.Activity=' . $row['Name'] . ' 
              AND registrations.Morning/Afternoon=' . $row['Morning/Afternoon'] . '';

然后

foreach($select as $select2)

此处$select是包含您的查询的字符串。通过对其进行操作，您将只是逐个获取字符串的所有字符，然后索引['activity']将不存在。

Answer 2

<?php

$sql = 'SELECT * FROM activities ORDER BY Id ASC';
if ($db->query($sql)) {
    foreach ($db->query($sql)->fetchAll(PDO::FETCH_ASSOC) as $row) {
        $select = 'SELECT COUNT(Activity) as Activity 
                   FROM `registrations` 
                   INNER JOIN activities 
                      ON registrations.Activity = activities.Name 
                   WHERE registrations.Activity=' . $row['Name'] . ' 
                      AND registrations.Morning/Afternoon=' . $row['Morning/Afternoon'] . '';
        if ($db->query($select)) {
        foreach ($db->query($select)->fetchAll(PDO::FETCH_ASSOC) as $select2) {
            $maxAantal = $row['Amount'] - $select2['Activity'];
        }

        echo '<tr>';
        echo '<td>' . $row['Name'] . '</td>';
        echo '<td>' . $row['Location'] . '</td>';
        echo '<td>' . $row['Begintime'] . '</td>';
        echo '<td>' . $row['Endtime'] . '</td>';
        echo '<td>' . $maxAmount . '</td>';
        echo '<td>' . $row['Deadline'] . '</td>';
        echo '<td>' . $row['Morning/Afternoon'] . '</td>';
        echo '<td>';
        echo '<a class="btn btn-secondary" href="readActi.php?id=' . $row['Id'] . '">Read</a>';
        echo '</td>';
        echo '</tr>';
        }
    }
}

$_SESSION['Id'] = $row['Id'];

?>

尝试考虑这段代码，在查询SQL语句后必须使用fetchAll才能使用foreach，而不再需要将其显式声明为数组。

由此：

foreach（$ db-＆gt; query（$ sql）as $ row）{

对此：

foreach（$ db-＆gt; query（$ sql） - ＆gt; fetchAll（PDO :: FETCH_ASSOC）为$ row）{

然后这个：

foreach（（array）$ select as $ select2）{

对此：

foreach（$ db-＆gt; query（$ select） - ＆gt; fetchAll（PDO :: FETCH_ASSOC）as $ select2）{

编辑：对于布尔问题，只需在每个声明中添加IF声明，请参阅上面的代码

SELECT maxAmount FROM活动 - SELECT COUNT（活动）FROM注册？

2 个答案: