[RESOLVED] Retrieving 'POST' data from Drop Down populated by database

nawrik

Hi,

I'm sure the solution will be straight forward - however I can't figure out where i'm going wrong.

I've successfully populated a drop down with MySQL table data:

$sql="SELECT id,gallerytitle FROM galleries ";

$result = mysql_query($sql);
?>
<form action="images_upload_single.php" method="POST">
<select name="users">
<option>Select Gallery...</option>
<?php
while($row = mysql_fetch_array($result))
  {
echo "

<option value =\"".$row['id']."\">".$row['gallerytitle']."</option>

";
};

?>
</select>
<input type="Submit" value="Save">
</form>

At the moment I just want to echo the correctly selected drop down option.

The following code doesn't seem to do this? any ideas what i'm missing out on here?

<?php
$galleryid=$_POST['id'];
$gallerytitle=$_POST['gallerytitle'];

echo $galleryid;
echo $gallerytitle;

?>

All help greatly appreciated!

DaiLaughing

If your print_r($_POST) it will give you a clue. It tells you what POST includes.

etully

Nice work so far.

You're very close.

The name inside the select field is the name of the variable that gets passed to your second page. So you wrote, "<select name="users">". So there is only one variable that's getting passed to the next page which you can retrieve as:

$_POST['users']

The value inside the quotes in the value="" tag is what gets passed through. Since you wrote: value =\"".$row['id']."\">

then $_POST['users'] is going to be in the form of the galleryid - that is:

$_POST['users'] is going to be an integer.

You could use that integer to look up the title in the database. That's the best way to handle this.

The other thing you could do (less optimal) is to change the value in the value="" tag to this:

<option value =\"".$row['gallerytitle']."\">".$row['gallerytitle']."</option>

That way, $_POST['users'] won't be an integer but it will be a string, it will be the actual title. I'm showing you this to illustrate how the value="" tag works. In practice, this is the wrong way to do it. You should normally pass the galleryid through.

nawrik

DaiLaughing, etully,

absolute legends!!!

As we all know - there is something really satisfying when you solve your own problem with PHP.

Without both your 'nudges' in the right direction I would still be pulling my hair out!!! THANKYOU.
I appreciated the 'test' - no better way to learn!

Firstly - I had never heard of or used

print_r($_POST)

When I submitted that, I realised the output was an integer...

I decided to do it the 'proper' way - so set about figuring out how to select what I want from my DB using that unique integer (the ID).

I know you guys know how to do it - but for anyone else searching - here is how I did it:
(I renamed 'users' to 'row')
My Form:

$sql="SELECT id,gallerytitle FROM galleries ";

$result = mysql_query($sql);
?>
<form action="images_upload_single.php" method="POST">
<select name="row">
<option>Select Gallery...</option>
<?php
while($row = mysql_fetch_array($result))
  {
echo "

<option value =\"".$row['id']."\">".$row['gallerytitle']."</option>

";
};

?>
</select>
<input type="Submit" value="Save">
</form>

When you hit 'Submit' it loads up "images_upload_single.php":

<?php

$selectgallery=$_POST['row'];

require_once("dbconnect.php"); // Include the file that connects to the database 

$query = "SELECT id,gallerytitle FROM galleries WHERE id='$selectgallery'";
$result = mysql_query($query) or die('Query failed: ' . mysql_error()); 

$galleryid = mysql_result ( $result, 0, "id" );
$gallerytitle = mysql_result ( $result, 0, "gallerytitle" );

echo $galleryid . $gallerytitle;


?>

This then successfully prints the selected items details from the table!

Thank you again!

DaiLaughing

Glad to help and also glad you took my limited help exactly the way I meant it.