Error message mysqli_fetch_array() expects parameter 1 to be mysqli_result

Question

1.00/5 (1 vote)

See more:

<pre><?php

session_start();
$con= mysqli_connect("localhost","root","","phplogin1");
$email = $_POST['email'];
$password = $_POST['password'];
$sql="SELECT * FROM 'accounts' where email='$email' and password ='$password'";

$result=mysqli_query($con,$sql);
$row = mysqli_fetch_array($result);


if ($row)
{
    
    header("location:home.php");
   $_SESSION['email'] = $_POST['email'];
   $_SESSION['loggedin'] = TRUE;
}
 else {
echo"Invalid";
//header ("refresh:2;url=login.php");
}
?>

What I have tried:

It is a very simple login page to for my application. I will develop the secure version later on. I having the error mysqli_fetch_array() expects parameter 1 to be mysqli_result when the user enter the correct email and password. The database will have the column username,password,email,image,address

Posted 7-Sep-20 19:41pm

ycwong99

Updated 7-Sep-20 21:39pm

v2

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Sandeep Mewara · Answer 1 · 2020-09-07T21:40:00

Quote:
mysqli_num_rows() expects parameter 1 to be in mysqli_result,bool given in

As per documentation of PHP: mysql_query - Manual[^] - when the query statement fails the call returns the boolean value FALSE.
It means, your query is returning FALSE and you expect a different type and have code based on it.

PHP

$row = mysqli_fetch_array($result);

Above, fetch query would not accept false as input parameter (which seems to be the value of $result). You need to act based on the $result value. Like:

PHP

$rows = mysqli_num_rows($query);
if($rows == 1){
    // login found
}
else
{
    // no user
}

OR

PHP

if($row){
            // login found
        }
    } else{
  // no user
}

Have a look at this to try and learn: How to Create A Simple Login Form in PHP & MySQLi Tutorial ~ Softaox[^]
Simple User Registration & Login Script in PHP and MySQLi | All PHP Tricks[^]