Question

我正在用PHP和SQLite创建一个登录系统，在将用户插入数据库的逻辑中，我的代码如下：

<?php

 class my_db extends SQLite3 {

    public $first_name;
    public $last_name;
    public $email;
    public $password;

    public function __construct($first_name, $last_name, $email, $password) {
        $this->first_name = $first_name;
        $this->last_name = $last_name;
        $this->email = $email;
        $this->password = $password;
        $this->open("didactic-eureka.db");
    }

    public function create_user_table() {
        $sql = <<<EOF
      CREATE TABLE IF NOT EXISTS USER(
          ID PRIMARY KEY NOT NULL ,
          FIRST_NAME TEXT NOT NULL ,
          LAST_NAME TEXT NOT NULL ,
          EMAIL TEXT NOT NULL ,
          PASSWORD TEXT NOT NULL 
      );
EOF;
        $this->exec($sql);
    }

    public function insert_user() {
        $sql = <<<EOF
     INSERT INTO USER(FIRST_NAME, LAST_NAME, EMAIL, PASSWORD)
     VALUES ($this->first_name,$this->last_name,$this->email,$this->password);
EOF;
46        $insert = $this->exec($sql);

        if (!$insert) {
            $this->lastErrorMsg();
        } else {
            echo "User Inserted into the database";
        }
    }
}


if (isset($_POST["submit"])) {
//    Get value from form
    $first_name = htmlspecialchars($_POST['first_name']);
    $last_name = htmlspecialchars($_POST['last_name']);
    $email = htmlspecialchars($_POST['user_email']);
    $password = htmlspecialchars($_POST['user_password']);

//    Check if values is empty
    if (empty($first_name) || empty($last_name)) {
        $err = "alert alert-danger";
        $err_msg = "Please enter a first or last name";
    } else {
        if (empty($email)) {
            $err = "alert alert-danger";
            $err_msg = "An email is required";
        } else {
            if (empty($password)) {
                $err = "alert alert-danger";
                $err_msg = "A password is required";
            } else {
                //    Hash password when all parameters are passed
                $hashed_password = password_hash($password, PASSWORD_DEFAULT);
                $db = new my_db($first_name, $last_name, $email, $hashed_password);

                // Create table
                $db->create_user_table();

                // Insert data into table
                $db->insert_user();
            }
        }
    }
}

?>

当我在浏览器中进行尝试时，它会返回以下响应：

警告：SQLite3 :: exec（）：“ @ gmail”附近：第46行上的/var/www/html/index.php中的语法错误

我该如何解决？

Answer 1

我终于找到了解决问题的办法。在我将从表单中获取的值插入数据库时，每个值都必须用单引号引起来，如下所示：

$sql = <<<EOF
INSERT INTO USER(FIRST_NAME, LAST_NAME, EMAIL, PASSWORD)
VALUES ('$this->first_name','$this->last_name','$this->email','$this->password');
EOF;

警告：SQLite3 :: exec（）：“。”附近：语法错误

1 个答案: