Pull down menu populated by database info - displaying selected option

bluedogsb · #1 (**permalink**) 06-24-09, 05:34 PM

So I have an html pull down menu populated by database information that is specific to a registered user. This makes it so only the user can choose his/her projects and no one else's. The trick is displaying the chosen project's information from the database on the same page. So basically if the user chooses a project from the menu, I need it to display that project and only that project.

The form code:

Code:

<form>
			<?
			include("mysqlconnect.php");
			
			$user = $session->username;
			$sql="SELECT projname FROM projects WHERE assignto = '$user' ";
			$result=mysql_query($sql);
			$options="";
				
				while ($row=mysql_fetch_array($result)) {
					$project=$row["projname"];
    				$options.="<OPTION VALUE=\"$project\">".$project.'</option>';
				}
				
			
			?>
			<select name="projname">
			<option>Choose Project
			<?=$options?>
			</option>
			</select>
			<input type="submit" value="Track Project">
			</form>

I realize I have no action in the form, but this is where you guys come in. I understand processing a form, but I can't get it to render on the same page. Email forms were easy, registration forms were easy but just saved the info. Can anyone do a quick help idea to help me render the data on the same page?

=OTS=G-Man · #2 (**permalink**) 06-25-09, 12:50 PM

well depands on whtat you want to do, if you want it to showup on the same page with out the page having to refresh, you could look into AJAX, otherwise just have your form submit back to itsefl using $_SERVER['PHP_SELF'] then at the top of your code just check to see if your form was submitted or not and it it was do your DB calls and get your info needed and display

so your code flow would look something line

PHP Code:


			
<?

if($_GET['projname'] != "")

{

    //get project info

    $content = display_project();

}

else

{

    $content = display_startpage();

}

?>



<html>

<body>

normal header code and project selector....

<?= $content ?>

normal footer stuff

</body>

</html>

hope it helps

bluedogsb · #3 (**permalink**) 06-26-09, 04:11 PM

I can't seem to get it to work for some reason, although different things have happened as I messed around with it. Is there a more specific explanation?

Jcbones · #4 (**permalink**) 06-26-09, 05:25 PM

What you are asking is really vague.

A simple answer to a vague question would be, use if/else statements.

PHP Code:


			

if(isset($_POST['projname']))
{

//Print this to the page... Everything that you want to show after the user chooses an option.

} else {

//Print the form, and dropdown options...

}

PS. It is bad practice to use short tags, as most servers have them turned off. Besides it is not hard to type that extra PHP...