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mysqli_query() expects parameter 1 to be mysqli, object given in C:\xampp\htdocs\Hospital management sysytem\logincode.php on line 10

What I have tried:


 while($row = mysqli_fetch_assoc($result)){
	$_SESSION['username'] = $row['username'];
	$_SESSION['password'] = $row['password'];
	header("SUCCESS LOGIN");
 echo "danger";

Updated 28-Oct-22 19:23pm
Richard Deeming 31-Oct-22 6:49am    
Based on that code snippet, you're storing your users' passwords in plain text. Don't do that!
Secure Password Authentication Explained Simply[^]

PHP even provides built-in functions to help you do the right thing:
PHP: password_hash[^]
PHP: password_verify[^]

1 solution

We can't tell: we have no access to your connection string, your code, or to your database. At a guess, $result is not what you think it is!

So try looking at this: PHP: mysqli_result::fetch_assoc - Manual[^] and compare the whole code with what you have written but not shown us.
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