警告：mysql_num_rows（）：提供的参数不是有效的MySQL结果资源

Question

Hi I'm receiving the error " Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/**/**/locate.php on line 16 ".

I've double checked everything, Googled/Stackoverflow searched and can't find out why its doing this. Would appreciate any thoughts!

getdate.php

function getDeals($the_type) {
$result = mysql_query("
    SELECT *
    FROM deals
    WHERE deal_type = '" . $the_type . "'
        ");
}

locate.php?type=fun

$type = $_GET['type'];
include("getdata.php");

getDeals($type);
if (mysql_num_rows($result)) {
    echo '<ul>';
    while($row = mysql_fetch_array($result))
        {
        echo '<a href="deal.php?i=' . $row["i"] . '">';
        echo '<li class="deal ' . $row["deal_type"] . 'deal">';
        echo '<h3>' . $row["deal_title"] . '</h3>';
        echo '</li>';
        echo '</a>';
        }
    echo '</ul>';
}
else {
    echo '<div class="nodeals">None</div>';
}

Answer 1

You're not returning the result from your getDeals function, so it isn't defined in the main body of the script.

function getDeals($the_type) {
    $result = mysql_query("SELECT * 
                             FROM deals
                            WHERE deal_type = '" . $the_type . "'");
    return $result;
} 

and

$result = getDeals($type); 

and ensure that your $the_type value is validated and escaped (or better yet, use PDO), to prevent SQL injection

Answer 2

You need to import the $result into the function as a global:

function getDeals($the_type) {
    global $result;
    $result = mysql_query("
        SELECT *
        FROM deals
        WHERE deal_type = '" . $the_type . "'
    ");
}

However, this is not a good way to set this type of thing up. You should instead use something like PDO or return a proper resource identifier.

Answer 3

There a number of ways to fix this, however the best way is to return the mysql resource and then assign that resource to the $result variable.

function getDeals($the_type) {
    return mysql_query("
        SELECT *
        FROM deals
        WHERE deal_type = '" . $the_type . "'
    ");
}

$result = getDeals($type);

Answer 4

As a rule of thumb, 99% of errors are due to given or the immediately previous line or statement. In your case, others noticed it first, that getDeals($type) does not return the result.

You should also change the call to mysql_query() to something like:

function getDeals($the_type) {
    $query = "
        SELECT *
        FROM deals
        WHERE deal_type = '" . $the_type . "'
            ";

    $result = mysql_query($query) or trigger_error(mysql_error().$query);

    return $result;
}