Selecting data via a dropdown box

[uwictech]

Hi all,

 

I'm having a brain freeze, hope you can help.

 

I have created a simple drop down box that populates from a MySQL database. When I choose one of the options from the drop down box, I would like to print on screen a value associated with it in the database. 

 

So when I choose a specific site, it will show the ip address. my code is below, hope you can help.

 

Jamie

 

 

<?php

	require("confwifi.php");

    
$extract = mysql_query ("SELECT * FROM site");
$numrows = mysql_num_rows ($extract);

echo"Select CAMPUS: <select name='site'>";

while ($row = mysql_fetch_assoc($extract))
{
$id  = $row['id'];
$site = $row['site'];
$ip = $row['ip'];

echo"<option name='$ip'>$site</option> ";


}

echo "</select>";


?>

<?php echo $ip; ?>

[taquitosensei]

what you're trying to do can't be done without javascript.    You would need to use javascript to submit the form onChange      then check for $_POST and do whatever you need to do.

 

      require("confwifi.php");

    
$extract = mysql_query ("SELECT * FROM site");
$numrows = mysql_num_rows ($extract);

echo"Select CAMPUS: <select name='site' onChange='this.form.submit();'>";

while ($row = mysql_fetch_assoc($extract))
{
   $id  = $row['id'];
   $site = $row['site'];
   $ip = $row['ip'];
   
echo"<option name='$ip'>$site</option> ";


}

echo "</select>";

echo ($_POST['site'])?$_POST['site']:"";
?>

[uwictech]

Thanks taquitosensei.

 

I don't think I explained myself properly.

 

What I would like is to do is, store what the user selects from the dropdown boxes into a variable to use later. Is that possible?

 

Sorry for the confusion.

 

Jamie

 

 

<?php

	require("confwifi.php");

$ip_form = $_POST['ip'];

$extract = mysql_query ("SELECT * FROM site");
$numrows = mysql_num_rows ($extract);

echo"Select CAMPUS: <select name='ip'>";

while ($row = mysql_fetch_assoc($extract))
{
$id  = $row['id'];
$site = $row['site'];
$ip = $row['ip'];

echo"<option name='$ip'>$site</option> ";



}

echo "</select>";


?>

<?php echo "$variable"; ?>

 

[Bricktop]

Hi Jamie,

 

I got your PM - sorry for the delay

 

OK, a simple bit of javascript should do what you need.  Add the following to the HTML of the page your PHP code runs from:

 

<script type="text/javascript">
function showSelected(val){
	document.getElementById
('selectedResult').innerHTML = "The IP address is - " 
+ val;
}
</script>

 

Now, amend your PHP code to read:

 

<?php

	require("confwifi.php");

$ip_form = $_POST['ip'];

$extract = mysql_query ("SELECT * FROM site");
$numrows = mysql_num_rows ($extract);

echo"Select CAMPUS: <select name='ip' onChange='showSelected(this.value)'>";

while ($row = mysql_fetch_assoc($extract))
{
$id  = $row['id'];
$site = $row['site'];
$ip = $row['ip'];

echo"<option name='$ip'>$site</option> ";



}

echo "</select>";

echo "<div id='selectedResult'></div>";

?>

<?php echo "$variable"; ?>

 

The above code will display the IP address when a user selects an option from the dropdown box.

 

Hope this helps.

[uwictech]

Hi Bricktop, there is most defiantly no need to apologise!

 

It's kind of doing what I want. When I choose an option from the drop down it echo's that out rather than the ip address associated with it the database.

 

Sorry to be a pain!

 

Jamie

 

[Bricktop]

Hi Jamie,

 

Just noticed you are using name= and not value= on your option boxes.  Change the relevant code to read:

 

echo"<option value='$ip'>$site</option> ";

 

Otherwise the code I posted won't work.

[uwictech]

Bricktop, you are indeed a genius!!

 

I hope someday I can help you in some way. I can't tell you how much I appreciate your help!

 

Jamie

[Bricktop]

No problem, if I'm ever in Wales I'll look you up - you can buy me a beer

 

Cheers.

[uwictech]

no problem there mate! I think I owe you more than 1!!

 

Can I ask one more question?

 

Is it possible to store that ip address in a variable, so that I can use it in some code I about to write?

 

J.

[Bricktop]

Hi Jamie,

 

I'm sure it's possible but beyond my scope of knowledge as it will need to be a relatively complex Javascript/AJAX call - because Javascript runs on the client and PHP runs on the server, and neither have direct access to each other.

 

I would recommend you post your code in the AJAX help forum and hopefully one of the experts there will give you an answer.

 

Thanks

[uwictech]

bugga!

 

I have just posted it onto the AJAX forum.

 

And there is me thinking it must be easy!

 

J.