Dropdowns!!

[7724]

Hi Guys,

Have created this menu:

<?php
//set up the query
$query = "select profiles.profilesid, profiles.secondname from profiles";
//run the query 
$result = mysql_query($query, $connection) or die(mysql_error());
//The following lines return the results from the query and assign them to variables
while($row = mysql_fetch_array($result)){
$secondname = $row['secondname']; 
$profilesid = $row['profilesid'];
//This line prints out the HTML Code with the returned data
echo "<option value='$profilesid'>$secondname</option>";
};
mysql_free_result($result);
?>

But I get this message:

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/sleaz/public_html/cms/add_report.php on line 1143

Now what I want it to do is get the profilesid and secondname from the profiles table and give me a list of those secondnames in the dropdown, but it's not working -any ideas?

Thanks

Chris

[chucklarge]

Firstly, Have you created a connection? where are you setting $connection ?

Secondly, your html form syntax for a dropdown is wrong, so you need to work on that, but that is not what is giving you the mysql error.

[7724]

Yeah, just did not include it in the example:

<?php
//connect to the DB 
include('connection.php');
//set up the query
$query = "select profiles.profilesid, profiles.secondname from profiles";
//run the query 
$result = mysql_query($query, $connection) or die(mysql_error());
//The following lines return the results from the query and assign them to variables
while($row = mysql_fetch_array($result)){
$secondname = $row['secondname']; 
$profilesid = $row['profilesid'];
//This line prints out the HTML Code with the returned data
echo "<option value='$profilesid'>$secondname</option>";
};
mysql_free_result($result);
?>

[Kudose]

First, unless you are connecting to multiple databases in your script, you don't need to add the $connection part to your mysql_query's. It will automatically know to use the DB your connected to. But you can leave it...no harm.

Second, in connection.php, add

 or die(mysql_error());

to your mysql_connect command. Or just show us the code, that would help more.

Then let us know what happens.

[Cailean]

well, the $connection variable is the [man]resource ink_identifier[/man] which is what's throwing the error...

connection.php should contain:

$connection=mysql_connect(...);
mysql_select_db(...);

if you are only using one db connection, you could leave the $connection variable out and mysql will connect to the last used connection.

edit: {cross-posted with above}