[RESOLVED] Dropdown with Submit to update mysql

[victimssun]

I am a beginner. I have a dropdown that is populated from a query from my db. I would like for the user to select something from the drop down, click submit and have this update a table. I have been able to find out how to update a table with the submit button if using text boxes, but not from a drop down. Here is what I have right now. I would like to send 'userID' (which is found based on the user logged in ->($query_row[userID])) and 'ingredientID' to the table 'useringredient'. I would assume I'd need an "insert.php" that would be part of this, but I don't know where to start with the form at. Any help would be appreciated. Thanks

<?php
// Connect to mysql
mysql_connect("*", "*", "*") or die(mysql_error());
mysql_select_db("*") or die(mysql_error());

// check cookies to make sure they are logged in
if(isset($_COOKIE['ID_my_site']))
{
   $username = $_COOKIE['ID_my_site'];
   $pass = $_COOKIE['Key_my_site'];
   $check = mysql_query("SELECT * FROM users WHERE username = '$username'")or die(mysql_error());
   $query = mysql_query("SELECT userID FROM users WHERE username = '$username'");
   $query_row = mysql_fetch_array($query);
   while($info = mysql_fetch_array( $check ))
   {

  // if the cookie has the wrong password, they are taken to the login page
  if ($pass != $info['password'])
  { header("Location: login.php");
  }

  // otherwise they are shown the member area
  else
  {
     // to make sure info is being pulled correctly
     echo "Welcome $username<p>";
     echo($query_row[userID]);

     // this is pulling the ingredients from db
     $result = @mysql_query("SELECT ingredientID,ingredient FROM ingredients");

     // this is putting the ingredients into a dropdown
     print "<p>Select ingredient(s):\n";
     print "<select name=\"ingredient\">\n";
     while ($row = mysql_fetch_assoc($result)){
        $ingredientID = $row['ingredientID'];
        $ingredient = $row['ingredient'];
        print "<option value=$ingredientID>$ingredient</option>\n";
     }
     print "</select>\n";
     print "</p>\n";

	 // link for drinklist
     echo "<a href=drinklist.php>View Current List</a>";
     print "</p>\n";
	 // logout
     echo "<a href=logout.php>Logout</a>";
  }
   }
}
else

//if the cookie does not exist, they are taken to the login screen
{
   header("Location: login.php");
}
?>

[CoderDan]

does this help any?:

<?php
// Connect to mysql
mysql_connect("*", "*", "*") or die(mysql_error());
mysql_select_db("*") or die(mysql_error());

// check cookies to make sure they are logged in
if(isset($_COOKIE['ID_my_site']))
{
	$username = $_COOKIE['ID_my_site'];
	$pass = $_COOKIE['Key_my_site'];
	$check = mysql_query("SELECT * FROM users WHERE username = '".mysql_real_escape_string( $username )."' ")or trigger_error(mysql_error(),E_USER_ERROR);
	if($info = mysql_fetch_assoc( $check ))
	{
		// if the cookie has the wrong password, they are taken to the login page
		if ($pass != $info['password'])
		{ header("Location: login.php");
		}

	// otherwise they are shown the member area
	else
	{
		// to make sure info is being pulled correctly
		echo "Welcome $username<p>";
		echo($query_row['userID']);

		// assuming the form posts to the same page, lets add our form handling code here:
		if ( isset( $_POST['ingredient'] ) )
		{
			$query = "SELECT * FROM ingredients WHERE ingerdientID = ".intval( $_POST['ingredient'] )." "; // assuming ingredientID is a unique/primary key so no limit needed
			$result = mysql_query( $query ) or trigger_error( mysql_error()."\nOriginal Query: $query", E_USER_ERROR );
			if ( $ingredient_array = mysql_fetch_assoc( $result ) )
			{
				// do whatever
				echo "<p>Found the ingredient you submitted: ".htmlentities( $ingredient['ingredient'] )."</p>\n";
			}
			else
			{
				echo "<p>ingredient not found</p>\n";
			}
		}
		// end form handling code

		// this is pulling the ingredients from db
		$result = mysql_query("SELECT ingredientID,ingredient FROM ingredients") or trigger_error(mysql_error(),E_USER_WARNING);

		// should start a form here
		echo "<form method='post' action='$_SERVER[PHP_SELF]'>\n";

		// this is putting the ingredients into a dropdown
		echo "<p><label for='ingredient'>Select ingredient(s):</label>\n";
		echo "<select id='ingredient' name='ingredient'>\n";
		while ($row = mysql_fetch_assoc($result)){
			echo "<option value='$row[ingredientID]'>".htmlentities( $row['ingredient'] )."</option>\n";
		}
		echo "</select>\n";
		echo "</p>\n";

		// should have a submit button and end of form here
		echo "<input type='submit' value='Submit Ingredient' />\n";
		echo "</form>\n";

	 // link for drinklist
		echo "<a href='drinklist.php'>View Current List</a>";
		echo "</p>\n";
	 // logout
		echo "<a href=logout.php>Logout</a>";
	}
}
else
{
	// no such user exists
}
}
else

//if the cookie does not exist, they are taken to the login screen
{
	header("Location: login.php");
}
?>

[victimssun]

This definitely helps. I'm now putting this project to the back of my list, but I understand how you're working things here. Thanks