Code Error - Please Help

[Anon]

I am getting an error on my site. It is Warning: Supplied argument is not a valid MySQL result resource in /www/eastcoastgolfconnection/getcities.php on line 74

Here is my code. What is wrong with it. Why is it not working.

						<?php
						$count=0; 
						$result=mysql_query("SELECT Distinct City from tblCourses where State = '$state' order by City"); 

						while ($r=mysql_fetch_array ($result) ) { 
						$City=$r["City"]; 
						$count++; 
						echo " <td>$City</td>"; 
						if ($count%3==0) //every third row 
						{ 
						echo "</tr><tr>"; 
						} 
						}  
						?>

[Anon]

You haven't connected to a MySQL server yet.

Diego

[Anon]

Yes I did. It is up at the top of my code. Does it need to be with the rest of the code? I am connecting.

[Anon]

hello jason,

the 'mysql_connect()' function returns
a connection-handle.
this you have to use with mysql_query() like:

$result=mysql_query($query,$conn_handle);

hth
chris

[Anon]

I don't understand. I am new to this. Here is all my code I have. What is wrong. I am soooo confused.

						<?php
						$link = mysql_connect("mysql.eastcoastgolfconnection.com", "eastcoastgolf", "golfanyone?");
						$query = "SELECT Distinct City from tblCourses where State = '$state' order by City";
						$result = mysql_query($query);

						$count=0;
						while ($r=mysql_fetch_array ($result) ) { 
						$City=$r["City"]; 
						$count++; 
						echo " <td>$City</td>"; 
						if ($count%3==0) //every third row 
						{ 
						echo "</tr><tr>"; 
						} 
						} 							
						?>

[Anon]

hello jason,

<?php
$link = mysql_connect("mysql.eastcoastgolfconnection.com", "eastcoastgolf", "golfanyone?");
$query = "SELECT Distinct City from tblCourses where State = '$state' order by City";
$result = mysql_query($query,$link);

                          $count=0; 
                          while ($r=mysql_fetch_array ($result) ) { 
                          $City=$r["City"]; 
                          $count++; 
                          echo " <td>$City</td>"; 
                          if ($count%3==0) //every third row 
                          { 
                          echo "</tr><tr>"; 
                          } 
                          } 
                          ?> 

hth
chris

[Anon]

Still not working

[Anon]

hello again,

first you might need to select the database
you want to use - anyhow i put some kind
of errorhandling on your code:

<?php
$which_db="YOUR_DATABASE";
if(!$link = mysql_connect("mysql.eastcoastgolfconnection.com", "eastcoastgolf", "golfanyone?"))
die("No Connect to DB");
if(!mysql_select_db($which_db,$link))
die("No Select DB");
$query = "SELECT Distinct City from tblCourses where State = '$state' order by City";
$result = mysql_query($query,$link);
if($result)
{
$count=0;
while ($r=mysql_fetch_array ($result) )
{
$City=$r["City"];
$count++;
echo " <td>$City</td>";
if ($count%3==0) //every third row
{
echo "</tr><tr>";
}
}
}
else
{
die("No Result with query: ".$query);
}
?>

hth
chris

[Anon]

Holy cow, It returned results. What did you do. Thank you sooo much. Would you mind if I contacted you on a needed basis. You are the sh*t. I have been working with this for days. I am a beginner and I really want to learn this language. Is there an email address or should I just keep posting here

[Anon]

hello again,

keep posting ;-)

bye
chris

[Anon]

Ok, I hope that you didn't post the actual username and password for your database server...coz if you did, change it now!!!

Diego