[SOLVED] i no this is a long shot but maybe

[jeppers]

hi again i am working on the same script but trying to update instead of selecting.

 

my problem is that i have 2x sql code and i am not sure if i am hedding down the correct road to victory.

 

the first set of sql code is trying to update and i feel like i am not in corporating the 2 tables need to be updated.

 

and the second is just trying to select the info to show the user. because there is two sets of code i am just not sure where the issues are:

// check the form has been submitted 
if (isset($_POST['submitted'])) {

//check for title
if (!empty($_POST['title'])) {
	$t = ($_POST['title']);
} else {
	$t = FALSE;
	echo '<p><font color="red">Please enter a food name.</font></p>';
}

//check for food description 
if (!empty($_POST['description'])) {
	$d =($_POST['description']);
} else {
	$d = FALSE;
	echo '<p><font color="red">Please enter a description of the food.</font></p>';
}

//check for food category 
if (isset($_POST['types']) && (is_array($_POST['types']))) {
	$type = TRUE;
} else {
	$type = FALSE;
	echo '<p><font color="red"> Please select one category.</font></p>';
}

//check food price
if (!empty($_POST['price'])) {
	$p = ($_POST['price']);
} else { 
	$p = FALSE;
	echo '<p><font color="red">Please enter a price</font></p>';
}

$result = mysql_query($query);
if (mysql_num_rows($result) == 0) {

// make query
$query = "UPDATE title='$t', description='$t', price='$p' AS f,
		  food_associations AS fa WHERE f.food_id = fa.food_id AND food_cat_id=$id";
		  $result = @mysql_query($query) or die(mysql_error());
		  if (mysql_affected_rows()==1) {// if ran ok 
		  
			//print message
			echo '<h1> Edit Food</h1>
			<p>The food has been updatted</p><p><br /><br /></p>';

		} else { //if it did not run ok
			echo '<h1> System error</h1>
			<p>Sorry there has been a system error and the food could not be updated. sorry</p>';
			echo '<p>' . mysql_error() .'<br /><br />Problem:' . $query . '</p>';// debugging message
			//include ('footer.php')
			exit();
		}

	} else { //report the errors
		echo '<h1>Error!</h1>
		<p>The follow errors have occurred:<br />';
		foreach ($errors as $msg){//print each message 
			echo "-$msg<br />\n";
		}
		echo'</p><p>Please try again.</p><p><br /></p>';
	}//end of if
}

//always show the form 

//retrive the users information 
$query = "SELECT f.food_id, title, description, price FROM foods AS f,
food_associations AS fa WHERE f.food_id = fa.food_id AND fa.food_cat_id=$type";
$result = @mysql_query($query) or die(mysql_error());

if (mysql_num_rows($result) == 1){//valid id show form 

	//get the users information
	$row = mysql_fetch_array($result, MYSQL_NUM);

 

please any advice will help

 

thanks