Dynamic <select>

stonebergftw · August 11, 2010

I'm having trouble getting the dynamic data from my <select> menus to write into my MYSQL database. Can anyone see what I'm doing wrong here? First post btw

The output looks like this, which is obviously wrong:

<html>
<head> </head>
<link rel="stylesheet" type="text/css" href="./css/newuser.css" />
<body>
<?php session_start();
require 'default.inc.php'; ?>

<?php if (isset($_POST['amount'])): 
	$host = 'localhost';
                $user = 'user';
                $pass = 'password';
                $conn = mysql_connect($host, $user, $pass);
                if (!$conn) {
                        exit('<p>Unable to connect to the database 
server</p>');
                }
                if (!@mysql_select_db('spikesusers')) {
                        exit('<p>Unable to locate the database</p>');
                }
        $locationname = $_POST['donor'];
        $donorid = mysql_query("SELECT id FROM donors WHERE locationname='$locationname'");

$amount = $_POST['amount'];
        $year = $_POST['year'];
        
$type = $_POST['type'];
        $typeid = mysql_query("SELECT id FROM donationtype WHERE type='$type'");

$player = $_POST['player'];
//$playerid = mysql_query("SELECT id FROM players WHERE 

        $sql = "INSERT INTO donations SET donorid='$donorid', amount='$amount', yearofdonation='$year', 
typeid='$typeid'";
mysql_query($sql);
?>
<div class='standard'>
<h1>Donation Management</h1>

<?php
if ($sql) {
                echo "New donation added ";
                echo "<p></p>";
	echo "<a href=managedonations.php>Back to donation management</a>";
                exit();
        } else {
        echo "Error adding new donation";
        echo "<a href=adddonation.php>Try again</a>";
        exit();
        }
        
?></div>
<?php else:
$host = 'localhost';
                $user = 'user';
                $pass = 'pass';
                $conn = mysql_connect($host, $user, $pass);
                if (!$conn) {
                        exit('<p>Unable to connect to the database
server</p>');
                }
                if (!@mysql_select_db('spikesusers')) {
                        exit('<p>Unable to locate the database</p>');
                }
$donor=@mysql_query('SELECT id, locationname FROM donors');
?>
<div class='standard'>
<h1>Donation Management</h1>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
    <label>Donor:
    <select class="text" name="donor">
<option value=new>Add a new donor...</option>
<?php
while ($donors=mysql_fetch_array($donor)) {
$donorname=$donors['locationname'];
echo "<option value='<?php echo $donorname;?>'>$donorname</option>";
//echo "<option value='hi'>hi</option>";
}
?>
</option>
</select>
</label><br />
<label>Amount:
      <input class="text" type="text" name="amount" class="text" /></label><br />
<label>Year of donation:
      <select class="text" name="year">
<option value='2011'>2011</option>
<option value='2010'>2010</option>
<option value='2009'>2009</option>
      </select>
</label><br />
<?php $player=@mysql_query('SELECT id, firstname, lastname FROM players'); ?>
<label>Player:
<select class="text" name="player">
<option value="player" selected="selected"></option>
<?php
while ($players=mysql_fetch_array($player)) {
$playerfirstname=$players['firstname'];
$playerlastname=$players['lastname'];
echo "<option value=player>$playerfirstname $playerlastname</option>";
}
?>
</select>
</label><br />

<?php $type=@mysql_query('SELECT id, type FROM donationtype'); ?>
<label>Donation type:
<select class="text" name="type">
<?php
while ($types=mysql_fetch_array($type)) {
$donationtype=$types['type'];
echo "<option value=type>$donationtype</option>";
}
?>
</select>
</label><br />

<input type="submit" value="SUBMIT" class="buttons"/>
    <input type="button" name="Cancel" value="CANCEL"
onclick="window.location = 'managedonations.php'" class="buttons"/>
</form>
</div> <?php endif; ?> </body>
</html>

Pikachu2000 · August 11, 2010

If you view the html source, you'll see that all of the <option>'s in the "donor", "year" and "player" <select> fields have the same value . . .

stonebergftw · August 12, 2010

Oops, that was a mistake, but...

Is there something wrong with the following code? As far as I understand, the value should change with every iteration of the while loop. Not working though :\

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
    <label>Donor:
    <select class="text" name="donor">
<option value=new>Add a new donor...</option>
<?php
while ($donors=mysql_fetch_array($donor)) {
$donorname=$donors['locationname'];
echo "<option value='<?php echo $donorname;?>'>$donorname</option>";
//echo "<option value='hi'>hi</option>";
}
?>
</option>
</select>

Pikachu2000 · August 12, 2010

A couple of things, don't use $_SERVER['PHP_SELF'] as a form action. To submit a form to itself, use action="" instead.

There's really no need to assign the result to another variable, you can just echo it directly.

You have an extra <?php echo in the <option value=

Also, there is an extra </option> closing tag at the end.

As long as the query is returning the expected results, your options should be populated.

Should end up looking like this:

<label>Donor:

<option value=new>Add a new donor...</option>

<?php

while ($donors=mysql_fetch_array($donor)) {

echo "<option value=\"{$donors['donorname']};\">{$donors['donorname']}</option>";

}

?>

</select>

stonebergftw · August 12, 2010

I fixed the problem with the following code:

<?php
$donor=@mysql_query('SELECT id, locationname FROM donors');

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
    <label>Donor:
    <select class="text" name="donor">
<option value=new>Add a new donor...</option>
<?php
while ($donors=mysql_fetch_array($donor)) {
$donorname=$donors['locationname'];
$donorid=mysql_query("SELECT id FROM donors WHERE locationname='$donorname'");
$donorid=mysql_fetch_array($donorid);
echo "<option value='$donorid[id]'>$donorname</option>";
}
?>

I think my code is pretty sloppy though. Oh well.

wildteen88 · August 12, 2010

There is no need for that query within your second while loop. You are already fetching the id and locationname from your original query (outside the loop)

$donor=@mysql_query('SELECT id, locationname FROM donors');

You can recode the while loop to just

<?php
while (list($id, $name) = mysql_fetch_row($donor))
{
     echo "<option value=\"$id\">$name</option>";
}
?>

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Dynamic <select>

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