mysql_query() returns returns true, but mysql_num_rows() and mysql_fetch_array() give "not a valid resource errors

Question

Here is the code in question:

From index.php:

require_once('includes/DbConnector.php');

// Create an object (instance) of the DbConnector
$connector = new DbConnector();

// Execute the query to retrieve articles
$query1 = "SELECT id, title FROM articles ORDER BY id DESC LIMIT 0,5";
$result = $connector->query($query1);

echo "vardump1:";
var_dump($result);
echo "\n";

/*(!line 17!)*/ echo "Number of rows in the result of the query:".mysql_num_rows($result)."\n";
// Get an array containing the results.
// Loop for each item in that array


while ($row = $connector->fetchArray($result)){

echo '<p> <a href="viewArticle.php?id='.$row['id'].'">';
echo $row['title'];
echo '</a> </p>';

From dbconnector.php:

$settings = SystemComponent::getSettings();

// Get the main settings from the array we just loaded
$host = $settings['dbhost'];
$db = $settings['dbname'];
$user = $settings['dbusername'];
$pass = $settings['dbpassword'];

// Connect to the database
$this->link = mysql_connect($host, $user, $pass);
mysql_select_db($db);
register_shutdown_function(array(&$this, 'close'));

} //end constructor

//*** Function: query, Purpose: Execute a database query ***
function query($query) {

echo "Query Statement: ".$query."\n";

$this->theQuery = $query;

return mysql_query($query, $this->link) or die(mysql_error());

}

//*** Function: fetchArray, Purpose: Get array of query results ***
function fetchArray($result) {

echo "<|";
var_dump($result);
echo "|> \n";

/*(!line 50!)*/$res= mysql_fetch_array($result) or die(mysql_error());

echo $res['id']."-".$res['title']."-".$res['imagelink']."-".$res['text'];

return $res;
}

Output:

Query Statement: SELECT id, title FROM articles ORDER BY id DESC LIMIT 0,5 vardump1:bool(true) 
PHP Error Message

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /*path to*/index.php on line 17

Number of rows in the result of the query: <|bool(true) |>

PHP Error Message

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /*path to*/DbConnector.php on line 50

possible duplicate of mysql_num_rows(): supplied argument is not a valid MySQL result resource — KingCrunch
– KingCrunch, Commented Aug 21, 2012 at 9:59

Ski · Accepted Answer · 2011-11-05 11:25:27Z

5

Your problem is in fallowing line:

return mysql_query($query, $this->link) or die(mysql_error())

it should have been written like this:

$result = mysql_query($query, $this->link);
if(!$result) die(mysql_error());
return $result;

It is common in dynamic languages that or returns first object which evaluates to true, but in PHP the result of X or Y is always true or false.

edited Nov 5, 2011 at 11:25

answered Jun 9, 2010 at 19:32

Ski

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1 Comment

zlance4012 Over a year ago

I just realized that when I echoed the mysql_error()... thanks people!

Your Common Sense · Accepted Answer · 2010-06-09 19:15:45Z

0

As you can learn from the manual, mysql_query return value type is not boolean but resource.
Something in your code converting it

answered Jun 9, 2010 at 19:15

Your Common Sense

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4 Comments

Dan Heberden Over a year ago

i bet the link is valid and his db_select is failing?

zlance4012 Over a year ago

mysql_db_select() gives bool(true) in var_dump, so it's not that.

zlance4012 Over a year ago

@ Col. Shrapnel - what could convert query result to a boolean? I posted code for all the $result variable lifetime. BTW, query statement works in phpmyadmin and returns valid results.

Your Common Sense Over a year ago

@zlance4012 nobody question your query statement. I am talking of mysql_query function result

Gary · Accepted Answer · 2010-06-09 19:28:55Z

0

Agree with above, you have an issue with your mysql_query. It should never return True. Either false or a resource.

Refactor : mysql_query($query, $this->link) or die(mysql_error())

echo mysqlerror() after your query and see what the error message is. You probably do not have a valid connection.

answered Jun 9, 2010 at 19:28

Gary

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2 Comments

Your Common Sense Over a year ago

@zlance4012 refactor it from silly childish die() to proper trigger_error()

Gary Over a year ago

For example : $dbconn = mysql_pconnect($myhost, $myuser, $mypass); if (!$dbconn) trigger_error('Database connection failure', E_USER_ERROR); When you use trigger_error, the error will be processed according to PHP's error handling routines.

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