PDO - 致命错误:在null上调用成员函数prepare()

时间:2016-12-14 10:18:02

标签: php mysql pdo

我在这里搜索了类似的问题,但无法解决我的问题。

我正在尝试向我的网站添加搜索功能,但由于某种原因搜索结果未显示。在我的error_log中,我收到此错误:

  

PHP致命错误:在第9行的/home/..../search.php中调用null上的成员函数prepare()

就是这一行:

<?php
require_once('includes/config.php'); 
include("includes/header.php");

// Search from MySQL database table
$search = $_POST['search'];
$query = $pdo->prepare("SELECT * FROM subs WHERE sub_title LIKE '%$search%' LIMIT 0, 10");
$query->bindValue(1, "%$search%", PDO::PARAM_STR);
$query->execute();

// Display search result 
if(!$query->rowCount() == 0) {
echo "Search found:<br>";
echo "<table>";
echo "<tr><td>Title</td><td>Category></td><td>Language</td><td>Time</td><td>Download</td></tr>";
while($results = $query->fetch()) {
    echo "<tr><td>";
    echo $results['sub_title'];
    echo "</td><td>";
    echo $results['category'];
    echo "</td><td>";
    echo $results['sub_lang'];
    echo "</td><td>";
    echo $results['timestamp'];
    echo "</td><td>";
    echo $results['sub_data'];
    echo "</td></tr>";
}

echo "</table>";
} else {
echo "Nothing found";
}

?>

这是我的search.php代码:

<?php 
$username = '------';
$password = '------';

try {
$conn = new PDO('mysql:host=localhost;dbname=-----', $username, $password, array(
PDO::ATTR_ERRMODE => PDO::ERRMODE_EXCEPTION
));
}

catch(PDOException $e) {
echo "Failed to connect to database!" . "<br><br>";
echo $e->getMessage();
}
?>

这是我的config.php

<form class="form-inline" action="search.php" method="POST">
     <input class="form-control" name="search" id="search" type="text" placeholder="Search">
 </form>

这是我在header.php

中的搜索表单
No such module MyLibrary

1 个答案:

答案 0 :(得分:0)

您正在使用:$query = $pdo->prepare

应该是:

$query = $conn->prepare()

因为我是一个懒惰的程序员,所以我创建了一个新类来节省一些时间:

class DBCommon
{
    private $conn;

    /** @var Common */
    public $common;

    public function __construct()
    {
        $database = new Database();
        $db = $database->dbConnection();
        $this->conn = $db;
    }

    public function runQuery($sql)
    {
        $stmt = $this->conn->prepare($sql);
        return $stmt;
    }
}