small mysql_num_rows error

Bastern

Hello and many thanks in advance for any help or advice. I have searched this forum for the topic - and indeed quite a few threads did come up - but they tended to be very specific with regards to toehr variables/jobs etc.

I am having a little trouble with a

$num_results = mysql_num_rows($results);

function.

When i tes this i get the following in my browser:

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\program files\apache group\apache\htdocs\phpwork\results.php on line 47

Without wishing to boh you down in the rest of my code or get boring with what i am trying to do, i shall breifly provide a little other php that i am using:

mysql_select_db('books');
   $query = "select * from books where ".$searchType." like '%".$searchTerm."%'";

   //run the query
   $result = mysql_query($query);

//tells me number of rows returned by the query
$num_results = mysql_num_rows($result);

   echo '<p>Number of books found: '.$num_results.'</p>';

The above is just to put my error in context if you see what i mean.

I would greatly apprecaite any help, so many thanks.

Thanks again.

Best regards, Bastern 🙂

LordShryku

Use this to get a more specific error

$result = mysql_query($query) or die(mysql_error());

Bastern

LordShryku,

Thank you very much for your quick response.

Sorry - i should have used the error() function to shed more light on the problem.

I used it and i see the following in my browser:

You have an error in your SQL syntax near 'like '%%'' at line 1

So naturally i ran to that linbe of code (and ones surrounding it) and double checked what was in the book, i could not see any difference.

The line of code causing an error is the following line in of code:

$query = "select * from books where ".$searchType." like '%".$searchTerm."%'";

Thanks for your time and help, i do appreciate it.

Oh - one thing i found odd was that it tells me that my error is on line 1 - but that specific line of code is on 41 - is it taking the start of that from the query, or starting from the SQL and ignoring the rest of the php/html. Thanks again in advance.

Regards, Bastern 🙂

Bastern

Sorry - having fiddled around, i have found another 'set-up' error.

I am not going to ask you about it, tell you about it or bore you with it.

Thanks, dont worry about a reply just yet.

Best regards, Bastern 🙂

TheIceman5

$searchTerm, what sets this value?

Bastern

TheIceman5,

Thanks for your help!
I set the values by using the following:

                $searchType=$HTTP_POST_VARS['searchType'];
	$searchTerm=$HTTP_POST_VARS['searchTerm'];
	$searchTerm=trim($searchTerm);

            $searchType = addslashes($searchType);
            $searchTerm = addslashes($searchTerm);

I am kind of new to php but not a complete newbie - i have created a message board in Flash with php and have been mucking around with it for a while, and i test everything on my localhost - which is why i was very surprised to discover that my browser is telling me that:

Access denied for user: '@' to database 'books'

But when i am in the MySQL monitor i am fine and it allows me to edit/delete and insert values into tables etc etc. What could i be doing wrong.

I only get the error when i am trying to select the database, not when i try to connect - so that is why i am stumped - and yes (no, honestly) - the database does exist !!!

Sorry for the rather idiotic quesitons.

Massive thanks for your time and effort.

Regards, Bastern 🙂

TheIceman5

are you entering anything in the search box?
if the query dont return nuthin then your gonna get errors when it tries to get the number of rows.
try

$num_results = @mysql_num_rows($results);

you wont get errors from it then, just null.

as for other one, what code ya using? i cant answer until i see.

Bastern

Thank you TheIceman5.

Thanks for your help. It turned out to be a problem with my query string - i dont know quuite how i fixed it to be honest, and as far as it telling my that i could not connect, well i just created a new pwd and username and that solved that.

I think it was a lethal combo of my own stupidity and two seperate problems occuring at once.

Anyway, cheers.

Regards, Bastern 🙂

laserlight

Generally a "supplied argument is not a valid MySQL result resource" message means that you should check the query involved, as it likely has a problem.