Jump to content

keep a variable from a mysql generated drop down


peehaichpee

Recommended Posts

hey all :)

so I have this bit down:

$query="SELECT `2010 Region Code` AS codes FROM locations";

$results = mysql_query($query);


$options="";

$options = "<select location='codes'>";
while($nt=mysql_fetch_assoc($results))
{

$thing=$nt["codes"];

$options.="\r\n<option value ='{$nt['codes']}'>
{$nt['codes']}</option>";
}
$options .="\r\n</select>";
echo $options;

 

what I'm trying to do is grab the selection from the drop down and display it as a table (the sql query would be extended should we manage to figure this one out

 

I've tried

echo"<form name='LOCATIONS' action='".$_SERVER['PHP_SELF']."' target='iframe' method='post'>";

any ideas?

Link to comment
Share on other sites

this is a javascript question, you dont need the form part

 

$options.="<option value =\"page.php?code={$nt['codes']}" onchange=\"window.open(this.options[this.selectedIndex].value,'myIFrame')\">{$nt['codes']}</option>\r\n";

 

Give the iframe the name myIframe, create another page in this example called page.php and setup your table, the variable $_GET['code'] in this example will be available on the page.php page for you to build your table from

Link to comment
Share on other sites

This thread is more than a year old. Please don't revive it unless you have something important to add.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Restore formatting

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.