PHP - MySQL Not connecting?

Question

Im trying to do a mysql query but for some reason this is the error I'm getting:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home1/server/public/chat/includes/functions/chat.func.php on line 9

Line 9 is as follows:

l6 $query = "SELECT `Sender`, `Message` FROM `database_chat`.`chat` ORDER BY `Msg_ID` DESC";    
l7 $run = mysql_query($query);
l8 $messages = array();
l9 while($message = mysql_fetch_assoc($run)) {...

database's name is 'database_chat' and 'chat' is the table...

For some reason it ain't connecting, what am I missing?

I've double-checked that the table elements are correctly written....

That means the query failed. Try executing it manually (using phpMyAdmin or command line) and see what you get? — Amal
– Amal, Commented Nov 6, 2013 at 17:04
mysql_* is deprecated, don't use it, but still if you want to solve your issue, try this: $run = mysql_query($query) or die(mysql_error()); — Horse SMith
– Horse SMith, Commented Nov 6, 2013 at 17:05
Please don't use mysql_* functions anymore, they are deprecated. See Why shouldn't I use mysql_* functions in PHP? for details. Instead you should learn about prepared statements and use either PDO or MySQLi. If you can't decide which, this article will help you. If you pick PDO, here is a good tutorial. — Marcel Korpel
– Marcel Korpel, Commented Nov 6, 2013 at 17:06
@AmalMurali the query that I ran on phpMyAdmin is good, everything is ok...Its the first time I try to run php and mysql on a hosting server, so Im getting the hang of this.... — user2766367
– user2766367, Commented Nov 6, 2013 at 17:07

Marcel Korpel · Accepted Answer · 2013-11-06 17:24:22Z

1

Try to use error checking in your code:

$con = mysql_connect(...);

if (!$con){
   // Error handling here
   print_r("SQL Error:". mysql_error());
   exit;
}

$run = mysql_query($query);

if (!$run){
   // Error handling here
   print_r("SQL Error:". mysql_error());
   exit;
}

This will only help you find the real problem you are having, which i am assuming is an error in your query.

Important Note: mysql_* functions are deprecated, use mysqli_* or PDO

edited Nov 6, 2013 at 17:24

Marcel Korpel

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answered Nov 6, 2013 at 17:06

Ibu

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Comments

Josh Harrison · Accepted Answer · 2013-11-06 17:06:53Z

0

Try this to see if there's a problem in your query itself:

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

Note that you should only use or die(mysql_error()); for debugging - remove it for production use.

Also, assuming you've already connected to the database, have you selected the database before you run the query?

mysql_select_db('database_chat');

answered Nov 6, 2013 at 17:06

Josh Harrison

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3 Comments

user2766367 Over a year ago

yup, I've selected the database and connected to it once I connect, I run the code I posted

user2766367 Over a year ago

Thank you very much for your answer, just showemd me what was wrong. Very helpful!

andrewsi Over a year ago

@user2766367 - if you've figured this out, you can add an answer with the solution in it, and accept it when you can.

bassxzero · Accepted Answer · 2013-11-06 17:10:59Z

0

mysql_query($query) requires a resource. An example is listed below using $link as the resource.

$link = mysql_connect($host, $user, $pass);
    if (!$link) {
        die('Could not connect: ' . mysql_error());
    }
    $db_selected = mysql_select_db($db, $link);
    if (!$db_selected) {
        die('Can\'t use database : ' . mysql_error());
    }

$results = mysql_query($query, $link);

answered Nov 6, 2013 at 17:10

bassxzero

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1 Comment

andrewsi Over a year ago

mysql_query does not need a resource. It's an optional second parameter - if it's not given, it will use the most recently opened connection as a default.

Vivek S · Accepted Answer · 2013-11-06 17:24:24Z

0

in your 17th line query requires a database connection.. follow this example..

    $con=mysqli_connect("localhost","usernamename","password","db_name");
$sql="SELECT * from db_table'";
$result = mysqli_query($con,$sql);
while($row = mysqli_fetch_array($result))
{

     <---- your code here ----->

     }

answered Nov 6, 2013 at 17:24

Vivek S

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1 Comment

Marcel Korpel Over a year ago

Why that apostrophe after db_table?

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