Possible Duplicate:
PHP Error: mysql_fetch_array() expects parameter 1 to be resource, boolean given
I'm very confused with this error, it shows when I try to return a result from the DB that doesn't exist ... I tried mysql_num_rows() but it returns the same error but instead of mysql_fetch_assoc expects ... it says mysql_num_rows() expects ...
I set error_reporting(0) to avoid showing this error, but I'm not satisfied with this solution ...
Here's the proper way to do things:
<?PHP
$sql = 'some query...';
$result = mysql_query($q);
if (! $result){
throw new My_Db_Exception('Database error: ' . mysql_error());
}
while($row = mysql_fetch_assoc($result)){
//handle rows.
}
Note the check on (! $result) -- if your $result is a boolean, it's certainly false, and it means there was a database error, meaning your query was probably bad.
You must check if result returned by mysql_query is false.
$r = mysql_qyery("...");
if ($r) {
mysql_fetch_assoc($r);
}
You can use @mysql_fetch_assoc($r) to avoid error displaying.
This is how you should be using mysql_fetch_assoc():
$result = mysql_query($query);
while ($row = mysql_fetch_assoc($result)) {
// Do stuff with $row
}
$result should be a resource. Even if the query returns no rows, $result is still a resource. The only time $result is a boolean value, is if there was an error when querying the database. In which case, you should find out what that error is by using mysql_error() and ensure that it can't happen. Then you don't have to hide from any errors.
You should always cover the base that errors may happen by doing:
if (!$result) {
die(mysql_error());
}
At least then you'll be more likely to actually fix the error, rather than leave the users with a glaring ugly error in their face.
The proper syntax is (in example):
$query = mysql_query('SELECT * FROM beer ORDER BY quality');
while($row = mysql_fetch_assoc($query)) $results[] = $row;
If you just want to suppress warnings from a function, you can add an @ sign in front:
<?php @function_that_i_dont_want_to_see_errors_from(parameters); ?>
You don't need to prevent this error message!
Error messages are your friends!
Without error message you'd never know what is happened.
It's all right! Any working code supposed to throw out error messages.
Though error messages needs proper handling. Usually you don't have to to take any special actions to avoid such an error messages. Just leave your code intact. But if you don't want this error message to be shown to the user, just turn it off. Not error message itself but daislaying it to the user.
ini_set('display_errors',0);
ini_set('log_errors',1);
or even better at .htaccess/php.ini level
And user will never see any error messages. While you will be able still see it in the error log.
Please note that error_reporting should be at max in both cases.
To prevent this message you can check mysql_query result and run fetch_assoc only on success.
But usually nobody uses it as it may require too many nested if's.
But there can be solution too - exceptions!
But it is still not necessary. You can leave your code as is, because it is supposed to work without errors when done.
Using return is another method to avoid nested error messages. Here is a snippet from my database handling function:
$res = mysql_query($query);
if (!$res) {
trigger_error("dbget: ".mysql_error()." in ".$query);
return false;
}
if (!mysql_num_rows($res)) return NULL;
//fetching goes here
//if there was no errors only
Source: Stackoverflow.com