mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource

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,

Answer 1

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

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

Sometimes suppressing the query as @mysql_query(your query);

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

Don't use the depricated mysql_* function (depricated in php 5.5 will be removed in php 7). and you can make this with mysqli or pdo

here is the complete select query

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        // code here 
    }
} else {
    echo "0 results";
}
$conn->close();
?>

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.

Answer 5

<?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."'");

Answer 6

First, check your connection to the database. Is it connected successfully or not?

If it's done, then after that I have written this code, and it works well:

if (isset($_GET['q1mrks']) && isset($_GET['marks']) && isset($_GET['qt1'])) {
    $Q1mrks = $_GET['q1mrks'];
    $marks = $_GET['marks'];
    $qt1 = $_GET['qt1'];

    $qtype_qry = mysql_query("
        SELECT *
        FROM s_questiontypes
        WHERE quetype_id = '$qt1'
    ");
    $row = mysql_fetch_assoc($qtype_qry);
    $qcode = $row['quetype_code'];

    $sq_qry = "
        SELECT *
        FROM s_question
        WHERE quetype_code = '$qcode'
        ORDER BY RAND() LIMIT $Q1mrks
    ";
    $sq_qry = mysql_query("
        SELECT *
        FROM s_question
        WHERE quetype_code = '$qcode'
        LIMIT $Q1mrks
    ");
    while ($qrow = mysql_fetch_array($sq_qry)) {
        $qm = $qrow['marks'] . "<br />";
        $total += $qm . "<br />";
    }
    echo $total . "/" . $marks;
}