使用PHP脚本更新MySQL表

Question

I am developing an android application, where I want to insert the address of the user to the database calling a PHP script on a website.

In the first try, it has to insert the address into the database and after that, it has update the same tuple.

This is the PHP script I have, but it gives an error in line number 36 (Call to a member function bind_param()).

Nevertheless, insertion is working perfectly fine.

class DbOperations1{

    private $con;
    function __construct(){

        require_once dirname(__FILE__).'/DbConnect.php';

    $db = new DbConnect();
    $this->con = $db->connect();
    }

    public function createUser ($name, $email, $password) {
        if($this->isUserExist($name, $email))
        {
            return 0;
        }else{
            $password = md5($password);
            $stmt = $this->con->prepare("INSERT INTO `test` (`id`,`name`, `email`, `password`) VALUES (NULL, ? , ? , ? );"); 
            $stmt->bind_param ("sss", $name, $email, $password);

            if ($stmt->execute()){
                return 1;
            } else {
                return 2;
            }
        }

    }

    public function Address($id_user, $address, $road, $city, $country) {
        if($this->isAddressExist($id_user, $address))
        {   
            $stmt = $this->con->prepare("UPDATE address a, users u SET `address`=$address,`road`=$road,`city`=$city,`country`=$country WHERE a.id_user=u.id"); 
            $stmt->bind_param("ssss", $address, $road, $city, $country);
            if ($stmt->execute()){
                return 2;
            } else {
                return 3;
            }
        }else{
            $stmt = $this->con->prepare("INSERT INTO `address` (`id_address`, `id_user`,`address`, `road`, `city`, `country`) VALUES (NULL, ?, ? , ? ,? ,? );"); 
            $stmt->bind_param ("sssss", $id_user, $address, $road, $city, $country);

            if ($stmt->execute()){
                return 0;
            } else {
                return 1;
            }
        }

    }

        public function userLogin($email, $password){
        $password = md5($password);
        $stmt = $this->con->prepare("SELECT id FROM users WHERE email = ? AND password = ?");
        $stmt->bind_param("ss",$email,$password);
        $stmt->execute();
        $stmt->store_result(); 
        return $stmt->num_rows > 0; 
        }

        public function getUserByemail($email){
        $stmt = $this->con->prepare("SELECT * FROM users WHERE email = ?");
        $stmt->bind_param("s",$email);
        $stmt->execute();
        return $stmt->get_result()->fetch_assoc();
    }

        private function isUserExist($name, $email){
        $stmt = $this->con->prepare("SELECT id FROM test WHERE name = ? OR email = ?");
        $stmt->bind_param("ss", $name, $email);
        $stmt->execute(); 
        $stmt->store_result(); 
        return $stmt->num_rows > 0; 
    }

        private function isAddressExist($id_user){
        $stmt = $this->con->prepare("SELECT id_address FROM address WHERE id_user = ?");
        $stmt->bind_param("s", $id_user);
        $stmt->execute(); 
        $stmt->store_result(); 
        return $stmt->num_rows > 0; 
    }


}

Answer 1

According to your comments, the error is because of call to prepare failing. Following code will allow to get more info:

        if($this->isAddressExist($id_user, $address))
        {   
            $stmt = $this->con->prepare("UPDATE address a, users u SET `address`=$address,`road`=$road,`city`=$city,`country`=$country WHERE a.id_user=u.id");
            if($stmt != False) {
              $stmt->bind_param("ssss", $address, $road, $city, $country);
              if ($stmt->execute()){
                  return 2;
              } else {
                  return 3;
              }
            } else {
              // this line will give an insight into an error message
              echo $this->con->error;
            }
        }

After getting the error message from MySQL:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'road=Sarak,city=Khost,country=Afghanistan WHERE a.id_user= 10' at line 1

we locate exact error location and there is $address variable used, while it shall be ? ! Such statement can't be prepared.

Using following update statement should fix it:

$stmt = $this->con->prepare("UPDATE address a, users u SET `address`=?,`road`=?,`city`=?,`country`=? WHERE a.id_user=u.id");

Answer 2

public function Address($id_user, $address, $road, $city, $country) {
        if($this->isAddressExist($id_user))
    {   
        $stmt = $this->con->prepare("UPDATE address SET `address`=?,`road`=?,`city`=?,`country`=? WHERE id_user= ?");
        if($stmt != False) {
          $stmt->bind_param("ssss", $address, $road, $city, $country,$id_user);
          if ($stmt->execute()){
              return 2;
          } else {
              return 3;
          }
        } else {
          // hopefully this line will give an insight into an error message
          echo $this->con->error;
        }
    } else{
            $stmt = $this->con->prepare("INSERT INTO `address` (`id_address`, `id_user`,`address`, `road`, `city`, `country`) VALUES (NULL, ?, ? , ? ,? ,? );"); 
            $stmt->bind_param ("sssss", $id_user, $address, $road, $city, $country);

            if ($stmt->execute()){
                return 0;
            } else {
                return 1;
            }
        }

    }

This should work. When it comes to parameter binding you have to use ? (question mark ) instead of the parameter name and bind the parameters by name in the correct order in bind_param function.

Answer 3

$stmt = $this->con->prepare("UPDATE address SET `address`=$address,`road`=$road,`city`=$city,`country`=$country WHERE a.id_user=u.id");

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