无法使用include从另一个php文件执行数据库语句
在将类define添加到StoreUser.php文件并从store.php文件调用它之后,我做错了什么,因为它不起任何作用?它在文件StoreUser.php中运行良好,无需类定义和函数(见下文)。无法使用include从另一个php文件执行数据库语句,php,Php,在将类define添加到StoreUser.php文件并从store.php文件调用它之后,我做错了什么,因为它不起任何作用?它在文件StoreUser.php中运行良好,无需类定义和函数(见下文)。 $dbhost = "localhost"; $dbname = "riskinarviointilomake"; $dbuser = "root"; $dbpass = ""; //private $email; if(isset($_GET["email"])){ $conn = new
$dbhost = "localhost";
$dbname = "riskinarviointilomake";
$dbuser = "root";
$dbpass = "";
//private $email;
if(isset($_GET["email"])){
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if (!( $stmt = $conn->prepare("INSERT INTO users(unique_id,email,
encrypted_password) VALUES(?,?,?)"))){
echo "Table creation failed: (" . $conn->errno . ") " . $conn->error;
}
$stmt->bind_param('sss', $unique_id,$email,$password);
$unique_id=123457764;
$password="df12467hh";
$result = $stmt->execute();
$stmt->close();
$conn->close();
}
?>
这不管用
class StoreUser{
$dbhost = "localhost";
$dbname = "riskinarviointilomake";
$dbuser = "root";
$dbpass = "";
//private $email;
public function store($email){
$conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if (!( $stmt = $conn->prepare("INSERT INTO users(unique_id,email, encrypted_password) VALUES(?,?,?)"))){
echo "Table creation failed: (" . $conn->errno . ") " . $conn->error;
}
$stmt->bind_param('sss', $unique_id,$email,$password);
$unique_id=123457764;
$password="df12467hh";
$result = $stmt->execute();
$stmt->close();
$conn->close();
}
}
您遇到了什么错误?没有。当我使用第二种情况时,MySQL数据库表中并没有添加任何行。第一种情况可以添加行。这是有点困难,因为这个linuxMint安装(灯)没有给出任何错误。我通过安卓截击(android-volley)得到了错误,但发现问题太神秘了。这就是为什么我只通过简单的例子在php代码上查找问题的原因。我找到了php.ini设置。我修改它以显示错误。现在我从php得到错误消息。存在语法和非对象问题。
<?php
include 'StoreTest.php';
if(isset($_GET['email'])){
$email = $_GET['email'];
$store = new StoreTest();
$store->store($email);
}
?>