我正在将一个json_encode对象返回给 ajax 调用。首先,有没有更好的方法来做到这一点?这是json_encode needed?
到我问题的根源。当我尝试将键设为变量时,它会引发“未定义错误”。这一行:var displayTriggers = trigger_rows;
有谁看到我做错了什么?
PHP:
try {
$con = getConfig('pdo');
$con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql_triggers = "
SELECT *
FROM triggers
";
$triggers_stmt = $con->prepare($sql_triggers);
$triggers_stmt->execute();
$triggers_rows = $triggers_stmt->fetchAll(PDO::FETCH_ASSOC);
$triggers_arr = array();
foreach ($triggers_rows as $triggers_row) {
$trigger_id = $triggers_row['id'];
$trigger_title = $triggers_row['trigger_name'];
$trigger_setting = $triggers_row['setting'];
$trigger_user = $triggers_row['user_id'];
$trigger_placement = $triggers_row['placement'];
$trigger_date = $triggers_row['date_changed'];
$trigger_active = ( $trigger_setting == '1' ) ? ' active' : '';
$html = '';
$html .= '<div class="triggerRow" data-placement="'.$trigger_placement.'">';
$html .= '<div class="triggerRowLeft">';
$html .= '<div class="triggerTitle">' . $trigger_title . '</div>';
$html .= '<div class="triggerText">' . $trigger_date . '</div>';
$html .= '<div class="triggerText">' . $trigger_user . '</div>';
$html .= '</div>';
$html .= '<div class="triggerRowRight">';
$html .= '<div class="triggerButton' . $trigger_active . '"></div>';
$html .= '</div>';
$html .= '</div>';
$data = array('html' => $html);
$triggers_arr[] = $data;
}
echo json_encode(['trigger_rows' => $triggers_arr]);
}
catch(PDOException $e) {
echo "Connection failed: " . $e->getMessage();
}
潇湘沐