mysql_fetch_array（）警告和mysql_num_rows（）警告[重复]

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• mysql_fetch_array()/mysql_fetch_assoc()/mysql_fetch_row()/mysql_num_rows etc… expects parameter 1 to be resource or result 32 answers
• mysql_fetch_array() expects parameter 1 to be resource, string given [duplicate] 3 answers

Im new (start today) with PHP and MySQL to one of my mini projects. i have my db and i make a PHP code to get user_id and score from the table but i get this errors and i don't know how to get ride of them: you can see in the end that im getting my requested response.

Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in /home/marathon/domains/marathon-il.com/public_html/Apps/million/view_all.php on line 15

Warning: mysql_num_rows() expects parameter 1 to be resource, null given in /home/marathon/domains/marathon-il.com/public_html/Apps/million/view_all.php on line 18 {"success":0,"message":"new_user"}

this is my getinfo.php:

<?php

$response = array();

require_once __DIR__ . '/db_connect.php';

$db = new DB_CONNECT();

$user = $_POST['userid'];

$result = mysql_fetch_array("SELECT * FROM users WHERE user_id = '$user'");

// check for empty result
if (mysql_num_rows($result) > 0){

        $userinfo = array();
        $userinfo["user_score"] = $result["score"];
        $userinfo["user_date"] = $result["date"];

        $response['info']= $userinfo;
    $response["success"] = 1;

    echo json_encode($response);;

} 
else {
    $response["success"] = 0;
    $response["message"] = "new_user";
    echo json_encode($response);
}
?>

please help to fix it - thanks.

</div>

Answer 1

you have to actually execute the query first. try looking up mysql_query

$result = mysql_query("SELECT * FROM users WHERE user_id = '$user'")
$ar = mysql_fetch_array($result);