[SOLVED] Trying to create drop down lists...

[Timb75]

that are populated using the database. I receive this error:

No database selected
SELECT Name FROM Teams

 

Name is a field inside the Teams table if this helps.

 

Here is the code I am using:

 

<?php 
$user = "";  
$host = ""; 
$password = ""; 
$dbName = "olevetpo_polls"; 

/* make connection to database */  
mysql_connect($host, $user, $password) OR DIE( "Unable to connect 
to database"); 
mysql_select_db($dbName); //did you forget this line? 

$sql = "SELECT Name FROM Teams"; 
$query = mysql_query($sql) or die(mysql_error() . "<br>$sql"); //use the or die(...) part ONLY IN DEVELOPMENT 
?> 

<form action="action" method="post">  
<select name="option">  

<?php  
while ($row = mysql_fetch_array($result)) {  
    echo "<option value=\"" . $row['Name'] . "\">" . $row['Name'] . "</option>\n";  
}  
?>  
</select>  
<input type="submit">  
</form>

 

I left the database name, but not the rest for obvious reasons.

 

First, I am a big time noob when it comes to php, I found this code here at phpfreaks. I do know that this only shows one drop list....

 

With that said, I need some other help too....

 

I am trying to create a college football human poll, so I need 24 more of these drop downs (stacked on top of each other and with #1, #2, #3, etc preceeding the dropdowns), but only one submit, after the final list.

 

How would I do this?

 

 

[rhodesa]

do you get any errors if you change this:

mysql_select_db($dbName); //did you forget this line? 

to

mysql_select_db($dbName) or die(mysql_error());

[steveh62]

this might help I was trying to make labels and checkboxes in a similar way

 

<?php

$labels = array(
            0 => '200mm x 200mm',
            1 => '200mm x 300mm',
            2 => '200mm x 400mm',
            3 => '200mm x 500mm',
            4 => '200mm x 600mm',
        );

foreach ($labels as $named1 => $label) {
    print '<label>' .$label. '</label><input id="sizes" name="sizes[' .$named1. ']" type="checkbox" value="' . $label. '" onclick="setChecks(this)"';
    //if ($key == 0){
 // print "checked";
 //}
    print ' /><input id="price" name="price[' .$named1. ']" type="text" class="dropdown" /><br/>';
}
?>

 

 

[Zane]

you have either mispelled the database name OR you are connecting to the wrong host.

Those are the only two problems I can think of that can occur for a mysql_select_db.

[Timb75]

do you get any errors if you change this:

mysql_select_db($dbName); //did you forget this line? 

to

mysql_select_db($dbName) or die(mysql_error());

 

That works....thanks.

 

Now for my nex question:

 

What do I need to do to create more than 1 drop down list that are stacked in different rows?

 

How would I create the table to do so? I know HTML, the PHP is what is kicking my butt.