Question

我想将数据插入到phpmyadmin表中，类似的代码在另一个页面中工作，但它在这个中不起作用。这是代码：

<?php
$hostname = 'localhost';
$username = 'root';
$password = '';
try {
    $conn = new PDO("mysql:host=$hostname;dbname=database", $username, $password);
    $conn->setAttribute(PDO::ATTR_ERRMODE,PDO::ERRMODE_EXCEPTION);
    echo "Connected successfully";
}
catch (PDOException $e) {

    echo $e->getMessage();

}   

if ( isset($_POST['NomPrenom']) && isset($_POST['date']) && isset($_POST['email']) && isset($_POST['description']) ) {
    $NomPrenom      = $_POST['NomPrenom'];
    $date           = $_POST['date'];
    $email          = $_POST['email'];
    $description    = $_POST['description'];

    try {

        $sql = "INSERT INTO demande (NomPrenom,date,email,description) 
                          VALUES ('$NomPrenom','$date','$email','$description')";
        $stmt = $conn->prepare($sql);
        if ($stmt->execute(array(
            $NomPrenom,
            $date,
            $email,
            $description
        ))) {

            echo "Data inserted";
        } else {

            echo "could not insert";
        }
    }
    catch (Exception $ex) {
        error_log($ex->getMessage());
    }
}
?>

当我提交页面时刷新没有错误，表格中没有添加任何内容。这是数据库中的表 table named demande

Answer 1

您的查询构造错误，您需要在要准备的查询中放置位置参数，然后在执行时传递这些参数的值。

$sql = "INSERT INTO demande (NomPrenom,date,email,description) 
                  VALUES (?,?,?,?)";
$stmt = $conn->prepare($sql);
if ($stmt->execute(array($NomPrenom,$date,$email,$description))) {
    echo "Data inserted";
} else {
    echo "could not insert";
}

或使用像这样的命名参数

$sql = "INSERT INTO demande (NomPrenom,date,email,description) 
                  VALUES (:nom, :dat, :em, :desc)";
$stmt = $conn->prepare($sql);
$params = array(':nom'=>$NomPrenom,
                ':dat'=>$date,
                ':em'=>$email,
                ':desc'=>$description);

if ( $stmt->execute($params) ) {
    echo "Data inserted";
} else {
    echo "could not insert";
}

将数据插入phpmyadmin表

1 个答案: