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

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PHP: Warning: sort() expects parameter 1 to be array, resource given

I get this error:

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

$sql = dbquery("SELECT * FROM `channels` WHERE `cat_slug` = ".$cat." ");
while($row = mysql_fetch_array($sql)){
$category =  $row["cat_name"];
$slug =  $row["cat_slug"];
// other
}

and $cat can be for example "funny"

how can i change the code to get it done?

dbquery is either a typo of the PEAR ->query method, or a copy/paste from the web of part of a script. Perhaps you forgot to include a file with this function? — Jim Wharton, Jun 28 '11 at 18:32

score 1 · Accepted Answer · answered Jun 28 '11 at 18:29

You're using PEAR so you'll want to set up the object first:

$db =& DB::connect('mysql://usr:pw@localhost/dbname');
if (PEAR::isError($db)) {
    die($db->getMessage());
}

then create a resource:

$res =& $db->query("SELECT * FROM `channels` WHERE `cat_slug` = '$cat'");

Then you can fetch an array if your fetchmod is set to ordered:

while ($res->fetchInto($row)) {
    echo $row[0] . "\n";
}

It looks like you were using a mixture of standard PHP and PEAR.

in standard PHP, you'd need to do it like this:

$sql = "SELECT * FROM channels WHERE cat_slug = '$cat'";
$res = mysql_query($sql, $conn) //where $conn is your db link stuff

THEN you could do a fetch array

yes the last one was the problem. It works with `$sql = dbquery("SELECT * FROM channels WHERE cat_slug = '$cat' ");` Thank you. — m3tsys, Jun 28 '11 at 18:49

score 0 · Answer 2 · answered Jun 28 '11 at 18:23

You have a syntax error in the query, and/or something else wrong with the database. Within your dbquery function, you'd need to have something like:

$result = mysql_query($sql) or (die(mysql_error());

which will abort the script and output the reason why the query failed.

However, given your query string, and your data going into it, the error is due to a lack of quotes around your $cat within the query:

SELECT ... WHERE `cat_slug`=funny;

Unless your table has a field called "funny", this is a syntax error. You need:

SELECT ... WHERE `cat_slug`='funny';

(note the quotes).

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

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