这是编辑后的代码?我仍然收到404错误,并且什么都没有发送到数据库表。提交表单后404页面上的网址后,我看到dbhh.php(以下文件)
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
// prepare sql and bind parameters
$stmt = $conn->prepare("INSERT INTO user_input(first_name1, last_name1, email1)
VALUES (:first_name1, :last_name1, :email1)");
// insert a row
$stmt->execute([
':first_name1' => $_POST["first_name1"],
':last_name1' => $_POST["last_name1"],
':email1' => $_POST["email1"]
]);
echo "New records created successfully";
}
catch(PDOException $e)
{
echo "Error: " . $e->getMessage();
}
$conn = null;
?>
答案 0 :(得分:1)
准备查询后,您已声明POST变量。首先,请确保将POST值分配给变量。
// insert a row
$first_name1 = $_POST["first_name1"];
$last_name1 = $_POST["last_name1"];
$email1 = $_POST["email1"];
$stmt = $conn->prepare("INSERT INTO user_input(first_name1, last_name1, email1)
VALUES (:first_name1, :last_name1, :email1)");
$stmt->bindParam(':first_name1', $first_name1);
$stmt->bindParam(':last_name1', $last_name1);
$stmt->bindParam(':email1', $email1);
$stmt->execute();
注意区别。我将POST放在查询之前。 bindParam
执行后,就可以获取值。
答案 1 :(得分:0)
什么@david said。尽管更紧凑的方法是跳过手动绑定,并立即按PDO::execute
传递参数:
$stmt->execute([
':first_name1' => $_POST["first_name1"],
':last_name1' => $_POST["last_name1"],
':email1' => $_POST["email1"]
]);
答案 2 :(得分:0)
尝试一下。希望对您有帮助。
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";
$conn = new mysqli($servername, $username, $password, $dbname);
$stmt = $mysqli->prepare("INSERT INTO user_input (first_name1, last_name1,email1) VALUES (?, ?)");
$stmt->bind_param("si", $_POST['first_name1'], $_POST['last_name1'],$_POST['email1']);
$stmt->execute();
echo "New records created successfully";
$stmt->close();
$conn->close();
?>