[Resolved] mysql/php : not a valid MySQL result resource

[phat-tramp]

Looking for some help from the experts:

Here is the code:

function getCouponRecord($recordID) {
global $default_dbname, $default_tablename;

echo ("getCouponRecord:recordID = $recordID<BR>");		
$link_id = db_connect();
mysql_select_db($default_dbname);
$query = "SELECT * FROM $default_tablename WHERE coupid = '$recordID'";
echo ("getCouponRecord:query = $query<BR>");	
$result = mysql_query($query) || error_message(sql_error());
if (!$result) {
    echo "Could not successfully run query ($query) from DB: " . mysql_error();
    exit;
}
echo ("getCouponRecord:result = $result<BR>");	
$currentCoupon = mysql_fetch_assoc($result);
echo ("getCouponRecord:currentCoupon.coupid = $currentCoupon['coupid']<BR>");
echo ("getCouponRecord:currentCoupon.couptext = $currentCoupon['couptext']<BR>");
echo ("getCouponRecord:currentCoupon.coupprinted = $currentCoupon['coupprinted']<BR>");		
mysql_close($link_id);

}

And here is my output:

getCouponRecord:recordID = 1
getCouponRecord:query = SELECT * FROM coupon WHERE coupid = '1'
getCouponRecord:result = 1

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in c:\inetpub\wwwroot\coupons\coup_common.inc.php on line 124
getCouponRecord:currentCoupon.coupid =
getCouponRecord:currentCoupon.couptext =
getCouponRecord:currentCoupon.coupprinted =

Is it telling me that $result is not a valid resource? The query isn't failing as far as I can tell...that's why I printed out the value of $result. I can run this query directly against my database.

What gives...I'm pulling my hair out on this one and I'm sure it's something stupid on my part.

TIA!

[aaron_karp]

Well, I confess that I didn't read every inch of your code very thoroughly, but was that $default_tablename variable in your $query a string? If so, mysql queried strings should be in quotes...

Therefore when you define the query variable it should look like this:

$query = "SELECT * FROM '$default_tablename' WHERE coupid = '$recordID'";

Let me know how it turns out...

[phat-tramp]

Thanks but that didn't work. It yielded a syntax error.

Anyway, I think the real question in my original post is this:

• why is $result from the query = 1
• is that an ok return value from mysql_query for a SELECT (the docs seem to think so)...there is plenty of error checking to ensure $result is valid before moving on.

Thanks for any help.

[dnszero]

Because $result is local to your getCouponRecord() function. You need to pass $result from that function back to the main program (or whatever called that function in the first place).

[dnszero]

Never mind, I just spotted the real problem!

Change this line:
$result = mysql_query($query) || error_message(sql_error());

to this:
$result = mysql_query($query) or error_message(sql_error());

The precedence of the || operator is causing it to evaulate the whole expression rather than short circuit has you intended. That's why $result == 1.

Sorry 'bout that miss diagnosis in the last post...need more coffee.

[phat-tramp]

Awesome!! That resolved the error and makes perfect sense.

Thank you!

My echo statements after the fetch_assoc still don't print anything, but at least now it's getting that far.

Anyone see anything else I'm doing wrong with the fetch and display? Otherwise I'll mark this resolved and get back to pulling my hair out...😉

Thanks for all your help! Just joined this community and it's the best/quickest responses I've seen.