Error Msg mysql_fetch_row(): ...

del753

Hi, im getting the following error in line 53:
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource

heres my query...
$sql="select count(*) AS count from shows1 where BOOKED LIKE ',x32,' AND ShowID = " . $_POST["id"];

$result=mysql_query($sql);
$row = mysql_fetch_row($result); <<<line 53
if ($row["count"] = 0 ){?>
<img src="images/free.gif" border="0">
<?}else{?>
<img src="images/booked.gif" border="0">
<?}?>

im stuck, help appreciated.

TheDefender

Change:

$sql="select count(*) AS count from shows1 where BOOKED LIKE ',x32,' AND ShowID = " . $_POST["id"];

$sql="select count(*) AS count from shows1 where BOOKED LIKE ',x32,' AND ShowID = '" . $_POST['id'] . "'";

bradgrafelman

The above solution posted by TheDefender will work just fine, because your error is most likely caused due to the fact that you didn't surround a string in quotes (' or ") in your query.

del753

thanks for the replys guys, tried that line and im getting the exact same error! same line 53 etc, any other thoughts?

bradgrafelman

Add this line RIGHT AFTER your mysql_query() call:

echo "Query was: " . $sql . "<br>Query successfull?: " . $result . "<br>"Number of rows matched: " . mysql_num_rows($result) . "<br>";

and paste the output here for us.

del753

did that, the error ..
Parse error: parse error, unexpected T_STRING, expecting ',' or ';' in /webtree/seat.php on line 53

bradgrafelman

For the love of God... sigh my mistake. thwacks self

echo "Query was: " . $sql . "<br>Query successfull?: " . $result . "<br>Number of rows matched: " . mysql_num_rows($result) . "<br>";

EDIT: Also noticed another error in your code:

if ($row["count"] = 0 ){?>

should be:

if ($row["count"] == 0 ) { ?>

"=" sets a variable's value, and "==" is a logical operator (the thing you want here).

EDIT2: I knew I wasn't crazy! :p Anyway, [man]mysql_fetch_row/man fetch's enumerated arrays, not associative arrays. If you want to use $row["count"], then change mysql_fetch_row/B to mysql_fetch_assoc/B !

del753

ha ha, ok with your new line..

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /webtree/seats.php on line 53
Query was: select count(*) AS count from shows1 where BOOKED LIKE ',x32,' AND ShowID = ''
Query successfull?:
Number of rows matched:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /webtree/seats.php on line 55

my page is loading and this error is kinda echoing in the middle of it...changed that line too, same error, thanks for your help

bradgrafelman

Why is the ShowID blank? Should it be? I doubt it... that's probably not returning any rows!

EDIT: After we figure this out, you need to also note my two edits from my previous post.

del753

"after we figure this out" i love your confidence!, i've been tearing my hair out for 6 hours....

anyhoo, yeah your right, the show id shouldnt be blank, you select one on the previous page, it does work & noted your 2 edits above. i dont know whats going on

the showid is in the ie address bar so it seems to be sending?

tekky

You should both read the Debugging 101 thread 😉

first of all your query should have been changed to THIS immediately after you found you had a problem, especially one dealing with a result set...

$result=mysql_query($sql) OR die('error querying "'. $sql . '" -- '.mysql_error());

that will give you extremely helpful information both on the error MySQL is giving you, PLUS on what exactly you are sending to the SQL server.

9/10 times the problem lies in the fact that your $sql var isnt what you expect it to be

[edit]
by the way bradgrafelman if you replace your resolved bit in your sig with my link, users will be able to click it and resolve their thread 😉
[/edit]

del753

thanks tekky, using your line i got the following:

error querying "select count(*) AS count from shows1 where BOOKED LIKE ',x32,' AND ShowID = ''" -- Unknown column 'BOOKED' in 'where clause'

as i am a 'newbie', i take this to mean i dont have a booked table in my db mabye? or am i way off?

bradgrafelman

D'oh! My bad, forgot about mysql_error().. I would have gotten there eventually 😉 Thanks tekky!

You are correct; there is no 'booked' collumn in MySQL. What's shows1's structure?

EDIT: Also, if you see the ShowID in the address bar, you are NOT posting ShowID, and need to change $POST['id'] to $GET['id'], though I wouldn't insert raw data from the address bar into a SQL query. Try...

if(!get_magic_quotes_gpc()) $ShowID = addslashes($_GET['id']); else $ShowID = $_GET['id'];

That way you're guarenteed that PHP is stripping quotes so as to prevent from SQL injections.

EDIT2: Code corrected; read next reply.

TheDefender

Make sure your variable name matches too. You started out with $POST['id'] and bradgrafelman has eluded to $GET['ShowID']... I guess my point is this: If you're passing the variable via the URL, thenjust make sure the ?id= part matches what your $GET['id'] index is. So if you're passing ?ShowID=, the it should be $GET['ShowID'], ,but if it's ?id=, then it should be $_GET['id']

bradgrafelman

My god I'm 0 for 2 tonight.

Sorry del; thank TheDefender for that save. If you see "?id=1234" in the URL, take all of what's been said above about $GET instead of $POST.

EDIT: Thanks for your patience through this.. it's been one of those nights/weeks/months/years/decades/lifetimes . 🙁

TheDefender

It's all good... I think we're all living that same life just at different times. 😉

del753

thanks for all your help guys, its almost 3 in the morning, so i'm gonna sleep on it and tackle it again tomorrow, because i'm hugely confused at this point!:queasy: