我是 PHP 新手,目前正在学习 PDO 并远离 MySQLi,我是 CMS 大学的学生,我正在学习这门类(class),老师希望我关联两个表并显示它们,他的 MySQLi 代码看起来像这样
$query = "SELECT * FROM categories WHERE cat_id = {$post_category_id}";
$select_categories_id = mysqli_query($connection,$query);
while($row = mysqli_fetch_assoc($select_categories_id)) {
$cat_id = $row['cat_id'];
$cat_title = $row['cat_title'];
echo "<td>{$cat_title}</td>";
}
我的 PDO 代码看起来像这样,它没有按应有的方式工作......
我之前在我的类(class)中做过几个 PDO 查询,但不知何故这不起作用。
$query = "SELECT * FROM categories WHERE cat_id = :post_category_id";
$result = $connection->prepare($query);
$result->execute(array(":post_category_id"=>$post_category_id));
while($row=$result->fetch(PDO::FETCH_ASSOC)){
extract($row);
$cat_id = $row['cat_id'];
$cat_title = $row['cat_title'];
echo "<td>$cat_title</td>";
}
我将不胜感激任何和所有的建议
最佳答案
如果您设置正确的连接数据库。它工作正常
$query = "SELECT * FROM categories WHERE cat_id = :post_category_id";
$result = $connection->prepare($query);
$result->bindParam(':post_category_id',$post_category_id, PDO::PARAM_INT);
$result->execute();
$result = $result->fetchAll(PDO::FETCH_ASSOC);
if ($result) {
foreach($result as $row){
$cat_id = $row['cat_id'];
$cat_title = $row['cat_title'];
echo "<td>".$cat_title."</td>";
}
}
关于php - PDO 查询关联表并显示结果,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47742693/