Help.

[Mindelude]

Seems that my site has been running really slow. What could be causing this? What I mean is for me, the site bearly won't load. Please list all possibilities. Thanks!~

[the_ancient]

what is your sites address?

[Mindelude]

www.mindelude.com

[Ron]

If you're patient enough, here is the error message you're getting:

b>Warning</b>:  mysql_connect(): Lost connection to MySQL server during query in <b>/home/YOURDIRECTORY/public_html/config.php</b> on line <b>29</b><br />
I cannot connect to the database because: Lost connection to MySQL server during query

Perhaps there's a problem with MySQL, perhaps you've got a bad query in there that's just eating up the server and hanging around until it times out. I don't know, but if the code is ok, you could open up a ticket to support with the type of error you're getting (and perhaps this snippet of an error message).

Good luck.

[JPC-Masood]

The problem is with your remote mysql server (at another hosting provider) which is not responding and appears to be down. You may like to move your database over to the server here. If you need help with transferring your database, please open a support ticket.

[Mindelude]

What do you mean? My hosting company? I'm with jaguar.

[Gwaihir]

If you need help with transferring your database, please open a support ticket.

What do you mean? My hosting company? I'm with jaguar.

Uh oh.. I think you'd best open a support ticket and have them guide you through it

[JPC-Masood]

What do you mean? My hosting company? I'm with jaguar.

Please check your config.php where you have setup remote mysql connection.

[Mindelude]

Oh, my bad. I'm sorry. I got the slow problem done but I still get an error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/jflashin/public_html/index.php on line 27

[jason]

You probably have a line of code somewhere in index.php that looks something like

$somevar = mysql_connect('server','user','pass');
mysql_select_db('database'); 

On line 27 you have a call to mysql_fetch_array(). Make sure that whatever comes before the = sign in the mysql_connect() line is between the parens after mysql_fetch_array.

... mysql_fetch_array($somevar) ... 

Without seeing your actual code I can't provide you with much better help than that.

--Jason

[Mindelude]

Thanks jason. On line 27 I get something like this.

while ($myrow = mysql_fetch_array($result))

[jason]

Sorry, my example from last night was incorrect. I forgot to include the actual query, which returns the $result variable. Somewhere before line 27 there should be a line that looks something like

$result = mysql_query('SELECT * FROM WHEREVER ...'); 

Right after that line try adding this code:

if(!$result) {
    die(mysql_error());
} 

It looks as though your query isn't running correctly and this code will help you understand why. What it will do is check that $result actually contains query result and, if it doesn't it will print an error message to the browser and then stop the script. Once you see that error message you should be able to get an idea of what's going wrong.

--Jason

[Mindelude]

Jason, thank you very much. I have a error that my developer will fix. Thank you!!