filling a drop down select box from MYSQL table

[stonebergftw]

Is there any way to fill a <select> box with entries from a mysql table, and then to have the entries POST to fill in another mysql table.

 

Example.

 

MYSQL table 1

id

name

1Tim

2Joe

3Bob

 

Have a drop down select box from table 1, and then fill in the selections in the auto-incrementing table 2, formatted as follows:

 

MYSQL table 2

id

name

 

Is this even possible?

[fenway]

Everything's possible -- but this is really two issues: one -- grab all of the values from all of the tables, and two -- deal with php to populate and write out html accordingly.

[stonebergftw]

Rules noted.

 

Ok, to specifics.  Can you explain why this is not working?  I think there is something wrong with the following:

echo "<option value='<?php echo $donorname;?>'>$donorname</option>";

 

Out of the following code:

 

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

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
    <label>Donor:
    <select name="donor">
<?php
while ($donors=mysql_fetch_array($donor)) {
$donorname=$donors['locationname'];
echo "<option value='<?php echo $donorname;?>'>$donorname</option>";
}
?>
</select>
<input type="submit" value="SUBMIT"/>
</form>

 

There could be something wrong with this code block, but I'm really not sure:

<?php
$locationname = $_POST['donor'];
$donorid = mysql_query("SELECT id FROM donors WHERE locationname='$locationname'");

$sql = "INSERT INTO donations SET donorid='$donorid'";
mysql_query($sql);
?>

 

The select list reads the data just fine, but the

INSERT INTO donations SET donorid='$donorid'"

is not executing properly.  Lets assume that all MYSQL tables are correct and the format of the tables is irrelevant.  Any ideas?  Help is really really appreciated.

[fenway]

Well, check mysql_errror() -- and remove the @ sign before mysql_query(), since that silences errors.

 

What does that echo return?