[RESOLVED] problem parsing MySql resource

[VadimK]

home someone can help me

DB_Connect();

$Query = sprintf("SELECT 
								`category_id`, 
								COUNT(*) 
							FROM 
								Businesses 
							WHERE 
								country = 'il' 
							GROUP BY 
								`category_id`
							", 
								mysql_real_escape_string($Country) 
							);

$Result = mysql_query($Query);
$Result = DB_Result_To_Array($Result);
return $Result;

and i received this error
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in ***/func_db.php on line 33
Array ( )

also tried parsing it as an array inside an array

		$res_array = array();
		for ($count=0;  $row = mysql_fetch_array($Result) ; $count++)
		{
			$res_array[$count] = $row;
		}
		return $res_array;

the query result should be

category_id	COUNT(*)
1	4
2	2
3	1

can't seem to resolve this issue

[dalecosp]

Count(*) will only get you a count of total rows in the DB, I'm pretty sure.

Of course, you should be checking the value of $Result before attempting any operation on it, anyways:

$Result = mysql_query($Query);

if (!$Result) {
   // some error handling here of your choice ... we'll go simple
   die("No result from DB query ... mysql said: ".mysql)error());
} else {
   // continue doing your thing ;-)

You should always have a test like this while in the debugging/coding phase of a project.

HTH,

[bradgrafelman]

Also note that this:

mysql_real_escape_string($Country)

is superfluous and might as well be removed since you never use that value in the SQL query.

[VadimK]

bradgrafelman;10991081 wrote:

Also note that this:

mysql_real_escape_string($Country)

is superfluous and might as well be removed since you never use that value in the SQL query.

it is like that because i started to bug test for any possible problems and hard coded the search term, obviously it will not be there in the working version

dalecosp: thanks a lot for the lines i will add them during the development process, unfortunately it will not solve my current predicament

is there another way to check the content of the object $Result that contains mysql_query($Query)
instead of using the mysql_fetch_array

[Bonesnap]

Make sure that your query is actually correct. As in, make sure tables/columns are spelled correctly, etc. Try running the query in phpMyAdmin or some other database application. If you really don't want to do that, make the following modification to see if there's a MySQL error:

$Result = mysql_query($Query) or die(mysql_error());

[bradgrafelman]

VadimK;10991088 wrote:

is there another way to check the content of the object $Result that contains mysql_query($Query)
instead of using the mysql_fetch_array

Well for one, $Result isn't an object; it's either going to be a boolean or a [man]resource[/man].

Otherwise, sure; there's [man]mysql_result/man or any of the various mysql_fetch_*() functions.

[VadimK]

Bonesnap;10991095 wrote:

Make sure that your query is actually correct. As in, make sure tables/columns are spelled correctly, etc. Try running the query in phpMyAdmin or some other database application. If you really don't want to do that, make the following modification to see if there's a MySQL error:

$Result = mysql_query($Query) or die(mysql_error());

thank you everybody this really solved my problem apparently i wasn't connecting with the right user which gave me bad response

never be sure when you copy your old project that everything will work right = LESSON LEARNED

Thank you