Php Drop Down Menu Problem

Cork_Skate

Hi there ..... having a bit of a problem, I have a drop down menu,
populated from my database and it works fine , see here below ....

<?php

$username = "username";
$password = "password";
$hostname = "localhost";	
$dbh = mysql_connect($hostname, $username, $password) 
	or die("Unable to connect to MySQL");
$selected = mysql_select_db("database",$dbh) 
	or die("Could not select my database");


function generate_box() { 

$sql = "SELECT ClubNo, ClubName FROM club"; 
$result = mysql_query($sql) or die(mysql_error());
$entries = mysql_num_rows($result); 
$options = '<option value="-1"></option>'; 

while($row = mysql_fetch_assoc($result)) { 
        $options .= "<option value=\"{$row['ClubNo']}\">{$row['ClubName']}</option>"; 
 } 

return $options; 

} 

$options = generate_box();

?>

And this is the HTML ..... when i used "Get" instead of "Post" it
passed the selection along with the URL so that works

<table>
<form method="post" name="forum_select" action="result.php"> <table cellspacing="0" cellpadding="0" border="0">

<tr><td border="1"><span>Select a club from the drop down list: <br>

<select name="by">

<? echo "$options"; ?>

<input type="submit" value="Submit"></span>

And heres the problem

i dont know if it is running the SELECT below, .... i dont know if $result gets the data from the query
and i need it to show those results (via the echo in the WHILE part) when displayed on result.php

I dont know how to tell it where on result.php it is meant to go either ??

something like <?php echo $_result; ?> i'd imagine but that is wrong i think ??

<?Php


if ($_REQUEST['submit'] == 'Submit') {
$username = "username";
$password = "password";
$hostname = "localhost";	
$dbh = mysql_connect($hostname, $username, $password) 
	or die("Unable to connect to MySQL");
$selected = mysql_select_db("database",$dbh) 
	or die("Could not select my database");
$clubID = $_POST['Clubno']; 
$result=mysql_query("SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, Groundno, YearEstablished, 

Status, ManagerNo, CoachNo FROM club WHERE Clubno=$clubID"); 
while ($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
echo "Club Name: {$row['ClubName']} <br> League: {$row['DomesticLeague']} <br> Telephone: {$row['TelePhoneNo']} <br> Fax No.: 

{$row['FaxNo']} <br> Email: {$row['EmailAddress']} <br> Website: {$row['Website']} <br> GroundNo: {$row['GroundNo']} <br> 

Established: {$row['YearEstablished']} <br> Status: {$row['Status']} <br> ManagerNo: {$row['ManagerNo']} <br> CoachNo: 

{$row['CoachNo']} <br>"; 
}
}

?>

Can someone help ..... but bare in mind .... i am new to Php !!

bpat1434

$ClubID = $_POST['Clubno'];

shoudl be:

$ClubID = $_POST['by'];

And, you need a closing </select> statement for the options.

~Brett

Cork_Skate

Originally posted by bpat1434
$ClubID = $_POST['Clubno'];

shoudl be:

$ClubID = $_POST['by'];

And, you need a closing </select> statement for the options.

I have changed that now ... but still have problems 11

bpat1434

I'm glad you have problems. What are they?

Also, modify your query to be:

"SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, 
Groundno, YearEstablished, Status, ManagerNo, CoachNo 
FROM club 
WHERE Clubno='$clubID'"

You should also add or die() statments.

So you could have:

<?php
$query = "SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, 
Groundno, YearEstablished, Status, ManagerNo, CoachNo 
FROM club 
WHERE Clubno='$clubID'";

$result = mysql_query($query) or die('<b>'.mysql_errno().'</b><br>'.mysql_error().'<br>');
?>

That way, if the query is WRONG, you will get an error with a line number AND what the error is. The line number will reference the line within the query (so line 1 is SELECT . . . line 2 is GroundNo, Year . . . etc).

~Brett

Cork_Skate

Here is my code now .....


<?php
$username = "username";
$password = "password";
$hostname = "localhost";	
$dbh = mysql_connect($hostname, $username, $password) or die("Unable to connect to MySQL");
$selected = mysql_select_db("database",$dbh) or die("Could not select my database");

$ClubID = $_POST['by']; 
$query = ("SELECT ClubName, DomesticLeague, TelePhoneNo, FaxNo, EmailAddress, Website, Groundno, YearEstablished, Status, ManagerNo, CoachNo FROM club WHERE ClubNo=$clubID");
echo "$query"; 

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

while ($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
extract($row);
echo "Club Name: $row[ClubName] <br> League: $row[DomesticLeague] <br> Telephone: $row[TelePhoneNo] <br> Fax No.: $row[FaxNo] <br> Email: $row[EmailAddress] <br> Website: $row[Website] <br> GroundNo: $row[GroundNo] <br> Established: $row[YearEstablished] <br> Status: $row[Status] <br> ManagerNo: $row[ManagerNo] <br> CoachNo: $row[CoachNo] <br>"; 
}

?>

What seems to be happening now is that there is a problem with the query .....

echo "$query";

Gives me ClubNo= at the end of the echo ....

and when i put it into '$clubID' .... i get ClubNo="

Any ideas on that ??

Cork_Skate

Hold up there people ..... Guess what it is .....you guessed a TYPE-OH

It is ClubID ...... not clubID

Ahh ..... eerrr ..... Sorry for being such a bother !!

bpat1434

Well OK then!!

~Brett

Cork_Skate

Originally posted by bpat1434
Well OK then!!

~Brett

Cheers for your help by the way !!

All is good now !!