Drop down menu not working

[ady01]

Im creating a drop down menu in PHP which pulls data from a table in MySqL : This is my code  >

 

<?php
require_once("config.php");

$db=mysql_connect($AddressBook_HOST,$AddressBook_Username,$AddressBook_Password);
mysql_select_db($AddressBook_DatabaseName,$db);

$query = "SELECT HouseNumber
          FROM Addresses
          ORDER BY HouseNumber";
$result = @mysql_query($query);
print "<p><select></p>";    
if (mysql_num_rows($result) > 0) :
while ($row = mysql_fetch_array($result)) :
	$HouseNumber = $row[HouseNumber];
	print "<p><option>value='$HouseNumber'>$HouseNumber</option></p>";
endwhile;
print "<p></select></p>";
endif;

?>

 

 

Its Kind of working but on the web page the option in the dropdown box is showing Value='$HouseNumber'>$HouseNumber  instead of the real data  ? Any Idears what i've done wrong here  ?

 

 

 

 

[DeFActo]

print '<p><option value="'$HouseNumber'">'$HouseNumber'</option></p>';

try that.

[ady01]

Better, but the dropdown box now just has $'HouseNumber' and under it it says "; endif; ?>

[DeFActo]

check this out http://www.phpfreaks.com/forums/index.php/topic,195463.0.html

[ady01]

Weird, still not getting a dropdown box now just getting the below in the space where the dropdown should be !!

 

 

'.$field.''; } ?>

 

 

[DeFActo]

<?php
require_once("config.php");

$db=mysql_connect($AddressBook_HOST,$AddressBook_Username,$AddressBook_Password);
mysql_select_db($AddressBook_DatabaseName,$db);

$query = "SELECT HouseNumber
          FROM Addresses
          ORDER BY HouseNumber";
$result = @mysql_query($query);
print "<p><select></p>";    
if (mysql_num_rows($result) > 0) {
while ($row = mysql_fetch_array($result)) {
	$HouseNumber = $row[HouseNumber];
	print '<p><option value="'.$HouseNumber.'">'.$HouseNumber.'</option></p>';
}
print "<p></select></p>";
}

?>

what about now?

found some mistake, sorry, try now.

[Xurion]

Why are you putting P tags around your other elements?

[ady01]

Still nothing, the attatched file is what the dropdown box looks like with this code mate  :  Just cant understand why the word 'housenumber' is appearing instead of the data !

 

 

 

 

 

[attachment deleted by admin]

[ady01]

Do you think I should try taking them all off  ? If im honest I put them there due to 'thinking' i needed them  ?

[ady01]

Think they need to be there taking them off just results in the below being shown on the page  >>>

 

"; if (mysql_num_rows($result) > 0) { while ($row = mysql_fetch_array($result)) { $HouseNumber = $row[HouseNumber]; print ''.$HouseNumber.''; endwhile} print ""; } ?>

 

 

[blueman378]

try something like this,

 

<form action="<?=$PHP_SELF?>" method="post">  <?php
global $database;
    $q = "SELECT HouseNumber
          FROM Addresses
          ORDER BY 'HouseNumber'
         ";   
    $result = $database->query($q) or die("Error: " . mysql_error()); 
    /* Error occurred, return given name by default */
    $num_rows = mysql_numrows($result);
    if( $num_rows == 0 ){
      return 'NONE FOUND!';
    }
?>

<select name="name">
<?php
   while( $row = mysql_fetch_assoc($result) ) {
    echo '<option value="' . $row[HouseNumber] . '">';
    echo $row[HouseNumber] . '</option>';
    }

?>
</select>
</form>

[ady01]

Well Thanks For the 'new way' not sure why all this is not working though still getting the below !!!

 

" method="post"> query($q) or die("Error: " . mysql_error()); $num_rows = mysql_numrows($result); if( $num_rows == 0 ){ return 'NONE FOUND!'; } ?>

 

I have done a whole search database here and cant get a simple drop down working !

[ady01]

*Bump*

[rarebit]

Back to your original version:

print "<select>";
while ($row = mysql_fetch_array($result)) :
	$HouseNumber = $row[HouseNumber];
print "<option>value='".$row['HouseNumber']."'>".$row['HouseNumber']."</option>";
}
print "</select>";

should work... however if this doesn't, could you show the html code which is outputted!

[ady01]

Sure im getting the below when run

 

 

Value='$HouseNumber'>$HouseNumber

 

 

 

I think I will stick with my original code as what blueman is suggesting just makes things worse.

[rarebit]

You should be getting something like:

<select>
<option>value='42'>42</option>
<option>value='69'>69</option>
</select>

oooh, I see it... it actually needs to look like this:

<select>
<option value='42'>42</option>
<option value='69'>69</option>
</select>

we're closing the tags too soon!

[ady01]

Good one ! Think im still missing a trick though as this is a screen shot of what i get (I have also reposted the code as it now stands!!)

 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>
<br />
<?php
require_once("config.php");

$db=mysql_connect($AddressBook_HOST,$AddressBook_Username,$AddressBook_Password);
mysql_select_db($AddressBook_DatabaseName,$db);

$query = "SELECT HouseNumber
          FROM Addresses
          ORDER BY HouseNumber";
$result = @mysql_query($query);
print "<p><select></p>";    
if (mysql_num_rows($result) > 0) :
while ($row = mysql_fetch_array($result)) :
	$HouseNumber = $row[HouseNumber];
	print "<p><option value='$HouseNumber'>$HouseNumber</option></p>";
endwhile;
print "<p></select></p>";
endif;

?>
</body>
</html>

 

[attachment deleted by admin]

[rarebit]

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>

<body>
<br />
<?php
require_once("config.php");

$db=mysql_connect($AddressBook_HOST,$AddressBook_Username,$AddressBook_Password);
mysql_select_db($AddressBook_DatabaseName,$db);

$query = "SELECT HouseNumber
          FROM Addresses
          ORDER BY HouseNumber";
$result = @mysql_query($query);
print "<p><select>";
if (mysql_num_rows($result) > 0)
{
while ($row = mysql_fetch_array($result))
{
	$HouseNumber = $row[HouseNumber];
	print "<option value='$HouseNumber'>$HouseNumber</option>";
}
print "</select></p>";
}

?>
</body>
</html>

[MattDunbar]

Have you tested your PHP installation on this server yet?

 

Make a page phpinfo.php with the contents:

<?php
phpinfo();
?>

If this code contains a large table of values its the code, if not you don't have PHP setup correctly which is what this looks like to me.

[ady01]

Yes everything else working fine just not this Dam drop down menu !!!

 

Sorry Rarebit, still same Problem !

[MattDunbar]

	print "<option value='$HouseNumber'>$HouseNumber</option>";

Replace that with:

	print '<option value="'.$HouseNumber.'">'.$HouseNumber.'</option>';

[ady01]

Still the same mate, im wondering if ORDERBY HouseNumber is the problem here, I know it shouldent be but something is telling me that i should maybe try ordering by ID or something (E.g taking data from HouseNumber and then order by ID number)

 

That make any sence  ?

[MattDunbar]

Well, then I guess you're going to need to get to more difficult debugging.  Try adding things such as echoing mysql_num_rows($result) check that it is grabbing the right data.  Tell me if that returns the correct number of rows.

[ady01]

*bump*