以下 PHP SQL 代码显示错误
可恢复的致命错误:类 PDOStatement 的对象无法转换为 /home/customer/xxxx/fetch_data.php 第 28 行中的字符串
我试图显示表格中的产品信息filter
那里有两个$statement->execute();
,一种是统计总搜索结果,另一种是显示当前页面的产品。
我是 PDO 方法的新手,并不是整体编码方面的专家。
The $filter_query = $stmt . 'LIMIT '.$start.', '.$limit.'';
正在引起问题。
通过该函数和相关函数,代码可以运行并显示数据。但如果我启用它,就会出现错误。
$limit = '5';
$page = 1;
if($_POST['page'] > 1)
{
$start = (($_POST['page'] - 1) * $limit);
$page = $_POST['page'];
}
else
{
$start = 0;
}
$search = "%samsung%";
$stmt = $connect->prepare("SELECT * FROM filter WHERE product_name LIKE :needle");
$stmt->bindParam(':needle', $search, PDO::PARAM_STR);
##
##
$filter_query = $stmt . 'LIMIT '.$start.', '.$limit.'';
$statement->execute();
$total_data = $stmt->rowCount();
$stmt = $connect->prepare($filter_query);
$stmt->execute();
$result = $stmt->fetchAll();
$total_filter_data = $stmt->rowCount();
$output = '
<h3> '.$total_data.' results found </h3>
and display each product
我按照建议尝试了以下代码Your Common Sense
分页和总搜索结果计数工作正常,但没有显示任何产品。
$limit = '5';
$page = 1;
if($_POST['page'] > 1)
{
$start = (($_POST['page'] - 1) * $limit);
$page = $_POST['page'];
}
else
{
$start = 0;
}
$name=str_replace(' ', '%', $_POST['query']);
$search = "%$name%";
$base_sql = "SELECT %s FROM filter WHERE product_name LIKE ?";
##
##
####
$count_sql = sprintf($base_sql, "count(*)");
$stmt = $connect->prepare($count_sql);
$stmt->execute([$search]);
$total_data = $stmt->fetchColumn();
####
$data_sql = $count_sql = sprintf($base_sql, "*")." LIMIT ?,?";
$stmt = $connect->prepare($data_sql);
$stmt->execute([$search, $start, $limit]);
$result = $stmt->fetchAll();
##
$output = '
<h3> '.$total_data.' results found </h3> ';
if($total_data > 0)
{
foreach($result as $row)
{
and display each product
使用以下行使代码显示一些数据,但分页不起作用。
$data_sql = sprintf($base_sql, "*");
$stmt = $connect->prepare($data_sql);
$stmt->execute([$search]);
$result = $stmt->fetchAll();