Login PHP and MySql generating error mysqli_fetch_array ()

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Login PHP and MySql generating error mysqli_fetch_array ()

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#1 by (2 votes)
#2 by (1 votes)

2

Good morning!

I have the following code to log in to the system:

<?php


require_once('conexao.php');

// FETCH DATA FROM FORM USING METHOD POST
// IF BUTTON NAME "LOGIN" IS SET
if (isset($_POST['login'])) {




// FETCH DATA FROM INPUT FIELD
$user = mysqli_real_escape_string($conexao, $_POST['usuario']);
$pass = mysqli_real_escape_string($conexao, $_POST['password']);

  // CHECK ALL FIELD HAS BEEN FILLED UP
 if ($user && $pass) {

   // QUERY FROM DATABASE
  $query= mysqli_query($conexao, "SELECT * FROM usuarios WHERE usuario='".$user."'");
  $checkuser= mysqli_num_rows($query);

   // CHECK IF USERNAME EXIST ON DATABASE
  if($checkuser != 1) {

    // I'LL BE SETTING A VARIABLE IF YOUR DOESN'T EXIST
   $error = "Username doesn't exist in our database!";
  }

   // FETCHING PASSWORD IN DATABASE WHERE USERNAME COINCIDES
  while ($row = mysqli_fetch_array($user)) {
   $checkpass= $row['senha'];


    // CHECK IF ENTERED PASSWORD MEETS THE USERNAME PASSWORD
   if ($pass== $checkpass) {

     // IF ALL OKAY SET SESSION
    setcookie("usuario", $user, time()+7200);
    $_SESSION['usuario'] = $user;
    $_SESSION['start'] = time();
    $_SESSION['expire'] = $_SESSION['start'] + (60 * 60 * 60);
    header("Location: ../admin.php");
    exit();
   } else {

     // SET VARIABLE THAT'LL SHOW IF USER PASSWORD IS INCORRECT
    $error = "Incorrect password!";
   }
  }
 } else {

  // SET VARIABLE IF ALL FIELD ARE NOT FILLED UP
 $error = "Please enter a username and password.";
 }
}


?>

When trying to login, in the "login.php" page, I get the following error:

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, string given in C:\xampp\htdocs\projetoqa\php\checklogin.php on line 32

php mysql

asked by anonymous 01.09.2015 / 16:21

2 answers

1

You need to pass $ query to mysqli_fetch_array instead of passing the variable $ user.

01.09.2015 / 16:26

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score 2 · Accepted Answer

Try this:

while ($row = mysqli_fetch_array($query))

The first parameter of mysqli_fetch_array() should be resulttype and not a string , which is the variable $user .

In your logic, put everything that comes after checking the user inside a else :

if($checkuser != 1) {
    // I'LL BE SETTING A VARIABLE IF YOUR DOESN'T EXIST
    $error = "Username doesn't exist in our database!";
}
else {
    $row = mysqli_fetch_array($query);
        ....

Otherwise this code will run when the user is not found and may result in an error. And since your query should return only one record, while is not required.