PHP - Sorting Dropdown List

Hi!   I have this problem that I have to solve in PHP. I have such a difficult with this programming and I would be so grateful if someone could help me solve this. The problem: A lecturer has a group of students who need to be assigned to one of two groups for the

Hello.  I am just trying to sort an array alphabetically.  Here's the code I'm using:


function GetAssocCharities($registrationNo)
{
$db = ConnectionString::ConnectTo_www_ago_mo_gov_data();
$sqlstring = "SELECT * FROM charityPartners where RegistrationNo LIKE '$registrationNo' ORDER BY Name";


$results = @mysql_query($sqlstring);

$org = null;


while ($resultArray = mysql_fetch_array($results))
{

$temp = new Registrants($resultArray['ARegNoFormatted'], $resultArray);

$org[] = $temp; 
}

return asort($org);

}


When I try to loop through the array using a foreach loop, I get an error that Invalid argument supplied for foreach().

Here's my foreach loop:

$lstCharities = OrgQueries::GetAssocCharities($registrant->RegNoFormatted);

foreach($lstCharities as $c)
{

//dom something here
}



I'm sure its something with not understanding list v array or something or that nature, but I'm not having much luck figuring it out with google.

Thanks!

Hi first tyvm for reading my help request me and my friend ar trying to build a order form (see png file ^^) so the main idea is some on go to order form.php   in that order form they fil in thear info like name, adress you know ^^   thean here is the treaky part at the and of the form they kan add products to there order by selecting it or search it from the drop down list the items in the dropdown list come from msql   the client can add more or delete items by clicking - or + whean the clint click send the client need to get a mail whit there a list of what theay have orderd and there unic order number wel so far we can do that but we wont to send another list to the company of order where the added products in the order ar sorted out based on the product id number   this is what we have so far http://clientconstru...sion.nl/valeri/    tyvm again voor anyone who read this ^^ Attached Files  bestel-form.png   37.78KB   0 downloads

Hi,

I am trying to call the data from Mysql but I am getting an empty drop down list, this is the code:


mysql:

Code: [Select]
create table years
(  yearID integer auto_increment,
   year varchar(30),
   primary key (yearID)
);

insert into years (yearID, year) values ('1', '2007-2008');
insert into years (yearID, year) values ('2', '2008-2009');
insert into years (yearID, year) values ('3', '2009-2010');
insert into years (yearID, year) values ('4', '2010-2011');
insert into years (yearID, year) values ('5', '2011-2012');
insert into years (yearID, year) values ('6', '2012-2013');

PHP:

Code: [Select]
<?php require_once('../Connections/connection.php'); ?>
<?php
 
$result = @mysql_query( "select yearID, year, from sss.years");

print "<p>Select a year:\n";
print "<select name=\"yearID\">\n";
while ($row = mysql_fetch_assoc($result)){
$yearID = $row[ 'yearID' ];
$year = $row[ 'year' ];
print "<option value=$yearID>$year\n";
}
print "</select>\n";
print "</p>\n";
?>

Thank you!

If I can get this fixed, I will have completed all but the admin login for this project - my first php/mysql project.

Here is what I need.  I have a list_records.php that list all the records in the table 'links' and the category each entry is in from the table 'categories'.

Here are my table structures.
Code: [Select]
-- Table structure for table `categories`
--

DROP TABLE IF EXISTS `categories`;
CREATE TABLE IF NOT EXISTS `categories` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `categories` varchar(37) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ;

-- --------------------------------------------------------

--
-- Table structure for table `links`
--

DROP TABLE IF EXISTS `links`;
CREATE TABLE IF NOT EXISTS `links` (
  `id` int(4) NOT NULL AUTO_INCREMENT,
  `catid` int(11) DEFAULT NULL,
  `name` varchar(255) NOT NULL DEFAULT '',
  `url` varchar(255) NOT NULL DEFAULT '',
  `content` varchar(255) NOT NULL DEFAULT '',
  PRIMARY KEY (`id`),
  KEY `catid` (`catid`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ;

On the update.php file, I have a form that lets me make changes to the record. 

Here is the codes for update.php

Code: [Select]
<? include "menu.php" ?>

<? include "db.php" ?>
<?php
$id=$_GET['id'];

$sql = "select * from links where id =$id";
$query = mysql_query($sql);
 
while ($row = mysql_fetch_array($query)){
 
    $id = $row['id'];
    $catid = $row['catid'];
    $name = $row['name'];
    $url = $row['url'];
    $content = $row['content'];
 
    //we will echo these into the proper fields
 
}
mysql_free_result($query);
?>
 
<table width="65%" align="center">
<tr><td align="left">

<form action="updated.php" method="post">

<input type="hidden" value="<?php echo $id; ?>" name="id"/>
 
<br>

 
<b>Website Name:</B><br> Change the name of the website listing.<br>
<input type="text" value="<?php echo $name; ?>" name="name"/>
 
<br>
 <br>
<b>URL:</b><br> Change the URL of the website listing.<br>
<input type="text" value="<?php echo $url; ?>" name="url"/>
 
<br>
<br>
 
<b>Description:</b><br> Change the description of the website listing.<br> Limit 255 characters.<br/>
<textarea name="content" cols="45" rows="4" wrap="soft"><?php echo($content);?></textarea>

<br>
 <?php
$result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); 
while ($row = mysql_fetch_array($result)) {
$id = $rows["id"]; 
$categories=$row["categories"];
$options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>';
};


?>


<SELECT NAME=catid>
<OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION>
</SELECT>

<?php
mysql_close();
?>

<div align="center">
<input type="submit" value="submit changes"/>
 </div>
</form>

<br>
</td></tr></table>


The part of he code I need help with is

Code: [Select]
<?php
$result = mysql_query("SELECT id, categories FROM categories") or die(mysql_error()); 
while ($row = mysql_fetch_array($result)) {
$id = $rows["id"]; 
$categories=$row["categories"];
$options.= '<option value="'.$row['id'].'">'.$row['id'].'-'.$row['categories'].'</option>';
};


?>


<SELECT NAME=catid>
<OPTION VALUE=selected><? echo $catid; ?><? echo $options; ?></OPTION>
</SELECT>

I want it to default to the category that the entry is in.  If you look, you will see in the select portion that I I have Code: [Select]
<$ echo $catid; ?> which echos the proper category ID, but if I use Code: [Select]
<? echo $categories; ?> it echos Writing, which is the last category in the list.  Yet, the $options echo the catid and it corresponding category.  How can I get the default option to echo BOTH the catid and category name while also listing all the other categories so that the records can be moved to a new category is needed?

Any help will be appreciated. Thank you in advance.