Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given..

[johnmerlino]

Hey all,

 

The mysql_query function is failing in the show function and I'm not sure why:

 

function index($item){
	db_connect();
	$query = "SELECT * FROM $item ORDER BY $item.id DESC";
	$result = mysql_query($query);
	$result = db_result_to_array($result);
	return $result;
}

function show($item, $id){
	db_connect();
	$query = sprintf("SELECT * FROM $item WHERE $item.id = '%s", 
		mysql_real_escape_string($id));
	$result = mysql_query($query);
	$row = mysql_fetch_array($result);
	return $row;
}

//Stuff that belongs in view
 $books = index('books');	

foreach($books as $book){
	echo $book['title'].'</ br>';
}

$book = show('books', 1);
echo $book['title'].'</ br>';

 

Thanks for any response.

 

[johnmerlino]

Here is the parameter I am passing into the show function:

$book = show('books', 1);
echo $book['title'].'</ br>';

[wildteen88]

This error usually occurs due to one of the following

• You're not connected to the database
  
• The query returned no results as there is no records with the id of 1
  
• Or your query failed due to an error.
  

 

To see if their is an error use the function mysql_error, eg

		$result = mysql_query($query) or trigger_error('MySQL Error: ' . mysql_error() . '<br />Query: '  . $query, E_USER_ERROR );

[johnmerlino]

The mysql_error() made it obvious that I was missing quotes around %s.

Thanks.