mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetc

Question

I am trying to select data from a MySQL table, but I get one of the following error messages:

mysql_fetch_array() expects parameter 1 to be resource, boolean given

or

mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given

or

Call to a member function fetch_array() on boolean / non-object

This is my code:

$username = $_POST['username'];
$password = $_POST['password'];

$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');

while($row = mysql_fetch_array($result)) {
    echo $row['FirstName'];
}

The same applies to code like

$result = mysqli_query($mysqli, 'SELECT ...');
// mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given
while( $row=mysqli_fetch_array($result) ) {
    ...

and

$result = $mysqli->query($mysqli, 'SELECT ...');
// Call to a member function fetch_assoc() on a non-object
while( $row=$result->fetch_assoc($result) ) {
    ...

and

$result = $pdo->query('SELECT ...', PDO::FETCH_ASSOC);
// Invalid argument supplied for foreach()
foreach( $result as $row ) {
    ...

and

$stmt = $mysqli->prepare('SELECT ...');
// Call to a member function bind_param() on a non-object
$stmt->bind_param(...);

and

$stmt = $pdo->prepare('SELECT ...');
// Call to a member function bindParam() on a non-object
$stmt->bindParam(...);

Answer 1

Your code should be something like this

$username = $_POST['username'];
$password = $_POST['password'];
$query = "SELECT * FROM Users WHERE UserName LIKE '$username'";
echo $query;
$result = mysql_query($query);

if($result === FALSE) {
    die(mysql_error("error message for the user")); 
}

while($row = mysql_fetch_array($result))
{
    echo $row['FirstName'];
}

Once done with that, you would get the query printed on the screen. Try this query on your server and see if it produces the desired results. Most of the times the error is in the query. Rest of the code is correct.

Answer 2

Make Sure You're Not Closing Database By using db_close() Before To Running Your Query:

If you're using multiple queries in a script even you're including other pages which contains queries or database connection, then it might be possible that at any place you use db_close() that would close your database connection so make sure you're not doing this mistake in your scripts.

Answer 3

Please check once the database selected are not because some times database is not selected

Check

mysql_select_db('database name ')or DIE('Database name is not available!');

before MySQL query and then go to next step

$result = mysql_query('SELECT * FROM Users WHERE UserName LIKE $username');

f($result === FALSE) {
    die(mysql_error());

Answer 4

If you tried everything here, and it does not work, you might want to check your MySQL database collation. Mine was set to to a Swedish collation. Then I changed it to utf8_general_ci and everything just clicked into gear.

I hope this helps someone.

Answer 5

Try the following code. It may work fine.

$username = $_POST['username'];
$password = $_POST['password'];
$result = mysql_query("SELECT * FROM Users WHERE UserName ='$username'");

while($row = mysql_fetch_array($result))
{
    echo $row['FirstName'];
}

Answer 6

<?php
    $username = $_POST['username'];
    $password = $_POST['password'];
    $result = mysql_query("SELECT * FROM Users WHERE UserName LIKE '".$username."'");

    while($row = mysql_fetch_array($result))
    {
        echo $row['FirstName'];
    }
?>

And if there is a user with a unique user name, you can use "=" for that. There is no need to like.

Your query will be:

mysql_query("SELECT * FROM Users WHERE UserName ='".$username."'");