MySQL fetch array expects parameter 1 to be resource

Question

<?php

include_once 'DB_Connect.php';

function getCategories() {
    $db = new DB_Connect();
    // array for json response
    $response = array();
    $response["Student"] = array();

    // Mysql select query
    $result = mysql_query("SELECT * FROM Student");

    while ($row = mysql_fetch_array($result)) {
        // temporary array to create single category
        $tmp = array();
        $tmp["id"] = $row["id"];
        $tmp["name"] = $row["name"];
        $tmp["number"] = $row["number"];
        $tmp["address"] = $row["address"];
        // push category to final json array
        array_push($response["Student"], $tmp);
    }

    // keeping response header to json
    header('Content-Type: application/json');

    // echoing json result
    echo json_encode($response);
}

getCategories();

?>

I have this API for my Android app, but I have a problem with this error below. Any idea out there?

<br />
<b>Warning</b>: mysql_fetch_array() expects parameter 1 to be resource, boolean given in <b>...\get_Student.php</b> on line <b>13</b><br />
{"Student":[]}

use mysql_error() after the query to see what exactly happened. — Abhik Chakraborty
– Abhik Chakraborty, Commented May 12, 2014 at 10:35
@AbhikChakraborty wait i research on how to use mysqlerror il be right back Suvitruf ,ubercooluk because i use it as API for android app — Giant
– Giant, Commented May 12, 2014 at 10:36
@HakHak $result = mysql_query("SELECT * FROM Student") or die(mysql_error()); — Latheesan
– Latheesan, Commented May 12, 2014 at 10:37

Latheesan · Accepted Answer · 2014-05-12 10:49:16Z

mysql_* is officially deprecated. Use either PDO or mysqli. This is how you do it with MySQLi library:

<?php

// Connect To Db
$mysqli = new mysqli("localhost", "my_user", "my_password", "my_database");
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

// Query Function
function getCategories()
{
    // Init
    global $mysqli;
    $response = array('Student' => array());

    // Query Db
    $query = "SELECT * FROM Student";
    if ($result = $mysqli->query($query)) {
        while ($row = $result->fetch_assoc()) {
            $response['Student'][] = array(
                'id'      => $row['id'],
                'name'    => $row['name'],
                'number'  => $row['number'],
                'address' => $row['address'],
            );
        }
        $result->free(); // free up memory
    }

    // Finished
    header('Content-Type: application/json');
    exit(json_encode(response));
}

// Test
getCategories();

// Close Db Connect
$mysqli->close();

You can have this code:

// Connect To Db
$mysqli = new mysqli("localhost", "my_user", "my_password", "my_database");
if (mysqli_connect_errno()) {
    printf("Connect failed: %s\n", mysqli_connect_error());
    exit();
}

In a file called db_connect.php and you can include it on the script you require db connection. Because the variable $mysqli is initiated outside of the function() { ... } scope; you need to use global $mysqli; to get access to it. This applies regardless of the connection code being on same file or being included from external file. Happy coding.

perfect i will study what you have written here so i wont commit the same mistake again thank you again sir happy coding

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