ch 3 Exercise 1

Phrog · October 23rd, 2012, 04:51 PM

I am working through exercise 1 at the end of chapter 3 and am having a problem. The book code is not in the book zip file so I entered the code and am getting the following error

Warning: mysql_fetch_array() expects parameter 1 to be resource, string given in C:\xampp\htdocs\ch3\ch3assign1.php on line 17

the header to the table displays properly but the mysql is not correct and I can't figure out where. Too be honest I'm not entirely sure how this works so it is difficult to find the error. Here is my code.

Code:

<?php

//connect to mysql
$connect = mysql_connect("localhost", "bp5am", "bp5ampass")
	or die("Check your server");
	
//connect to database
mysql_select_db("moviesite")
	or die(mysql_error());
	
//function to get lead actor
function get_leadactor($lead_actor) {
	global $actorname;
	$query2 = "SELECT people_fullname " .
			"FROM people " .
			"WHERE people.people_id = $lead_actor ";
	$results = mysql_fetch_array($query2)
		or die(mysql_error());
	extract ($rows);
	$actorname = $people_fullname;
	}
	
//function to get director
function get_director($director) {
	global $directorname;
	$query2 = "SELECT people_fullname " .
			"FROM people " .
			"WHERE people.people_id = $director ".
	$results = mysql_fetch_array($query2)
		or die(mysql_error());
	extract ($rows);
	$directorname = $people_fullname;
	}

//create the table
//format a header row
echo "<table border = '1'>\n";
	echo "<tr>\n";
		echo "<td><strong>Movie Name</strong></td>\n";
		echo "<td><strong>Lead Actor</strong></td>\n";
		echo "<td><strong>Director</strong></td>\n";
	echo "</tr>\n";
		
//get the movies
$query = "SELECT * FROM movie ";
$results = mysql_query($query)
	or die (mysql_error());
	
while ($rows = mysql_fetch_assoc($results)) {
	extract ($rows);
	
	//run functions
	get_leadactor($movie_leadactor);
	get_director($movie_director);
	
	//build the rest of the table
	echo "<tr>\n";
		echo "<td>\n";
		echo "$moviename";
		echo "</td>\n";
		echo "<td>\n";
		echo "$actorname";
		echo "</td>\n";
		echo "<td>\n";
		echo "$directorname";
		echo "</td>\n";
	echo "</tr>\n";
	}
	
echo "</table>\n";

?>

UseLess · October 24th, 2012, 04:24 AM

Greetings,

In the 2 functions you setup a query but do not execute it, so the function 'get_leadactor' should be like this;

Code:

//function to get lead actor
function get_leadactor($lead_actor)
{
	global $actorname;

	$query2 = "SELECT people_fullname " .
			"FROM people " .
			"WHERE people.people_id = $lead_actor ";
	$results = mysql_query($query2) or die (mysql_error());

	$rows = mysql_fetch_array($results) or die(mysql_error());

	extract ($rows);

	$actorname = $people_fullname;
}

Phrog · October 24th, 2012, 09:57 AM

Okay, I did the change (thanks by the way) but I am still getting an error.

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\ch3\ch3assign1.php on line 33
Query was empty

the new code is

Code:

<?php

//connect to mysql
$connect = mysql_connect("localhost", "bp5am", "bp5ampass")
	or die("Check your server");
	
//connect to database
mysql_select_db("moviesite")
	or die(mysql_error());
	
//function to get lead actor
function get_leadactor($lead_actor) {
	global $actorname;
	$query2 = "SELECT people_fullname " .
			"FROM people " .
			"WHERE people.people_id = $lead_actor ";
	$results = mysql_query($query2)
		or die(mysql_error());
	$rows = mysql_fetch_array($results)
		or die(mysql_error());
	extract($rows);
	$actorname = $people_fullname;
	}
	
//function to get director
function get_director($director) {
	global $directorname;
	$query2 = "SELECT people_fullname " .
			"FROM people " .
			"WHERE people.people_id = $director ".
	$results = mysql_query($query2)
		or die(mysql_error());
	$rows = mysql_fetch_array($results)
		or die(mysql_error());
	extract($rows);
	$directorname = $people_fullname;
	}

//create the table
//format a header row
echo "<table border = '1'>\n";
	echo "<tr>\n";
		echo "<td><strong>Movie Name</strong></td>\n";
		echo "<td><strong>Lead Actor</strong></td>\n";
		echo "<td><strong>Director</strong></td>\n";
	echo "</tr>\n";
		
//get the movies
$query = "SELECT * FROM movie";
$results = mysql_query($query)
	or die (mysql_error());
	
while ($rows = mysql_fetch_assoc($results)) {
	extract ($rows);
	
	//run functions
	get_leadactor($movie_leadactor);
	get_director($movie_director);
	
	//build the rest of the table
	echo "<tr>\n";
		echo "<td>\n";
		echo "$movie_name";
		echo "</td>\n";
		echo "<td>\n";
		echo "$actorname";
		echo "</td>\n";
		echo "<td>\n";
		echo "$directorname";
		echo "</td>\n";
	echo "</tr>\n";
	}
	
echo "</table>\n";

?>

UseLess · October 24th, 2012, 10:47 AM

Greetings,

You have a . instead of a ; at the end of the query to grab the directors name.

Phrog · October 24th, 2012, 12:06 PM

That worked! Thanks so much, the devil is always in the details ...

Phrog