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Need Help with converting an ID to a Name

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Join Date: Oct 2007

Posts: 12

Reputation: jtmcgee is an unknown quantity at this point

Solved Threads: 0

jtmcgee

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Newbie Poster

Need Help with converting an ID to a Name

Oct 7th, 2007

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
	if( $rand_num <= $chance )
		{
			//$item_id = mysql_query("SELECT `item` FROM `creatures` WHERE id = '$id_creature'");
			$foo =& creature_stats( item, $id_creature );
 
			$item_query = mysql_query("SELECT * FROM items WHERE id = '$foo'") or ('$item_query');
 
			$item = mysql_fetch_array($item_query);
 
			echo"You got an ".$item."!";
 
		}
	if( $rand_num <= $chance )
		{
			//$item_id = mysql_query("SELECT `item` FROM `creatures` WHERE id = '$id_creature'");
			$foo =& creature_stats( item, $id_creature );
			
			$item_query = mysql_query("SELECT * FROM items WHERE id = '$foo'") or ('$item_query');

			$item = mysql_fetch_array($item_query);

			echo"You got an ".$item."!";

		}

Creature Stats Function:

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
function &creature_stats($what_stat, $id) 
{
$mysql_query=mysql_query("SELECT * FROM `creatures` WHERE `id` = '$id'");
$info=mysql_fetch_array($mysql_query);
return $info[$what_stat];
}
function &creature_stats($what_stat, $id) 
{
$mysql_query=mysql_query("SELECT * FROM `creatures` WHERE `id` = '$id'");
$info=mysql_fetch_array($mysql_query);
return $info[$what_stat];
}

I need help getting it to echo $item

It justs puts out nothing.

Last edited by jtmcgee; Oct 7th, 2007 at 5:50 pm.

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Join Date: Jun 2007

Posts: 1,227

Reputation: kkeith29 has a spectacular aura about

Solved Threads: 167

kkeith29

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Nearly a Posting Virtuoso

Re: Need Help with converting an ID to a Name

Oct 7th, 2007

try this, it might work.

 Help with Code Tags 
 PHP Syntax (Toggle Plain Text) 
if( $rand_num <= $chance )
		{
			$item_id = mysql_query("SELECT `item` FROM `creatures` WHERE id = '$id_creature'");
			$foo =& creature_stats( item, $id_creature );
 
			$item_query = mysql_query("SELECT * FROM items WHERE id = '$foo'") or ('$item_query');
 
			$item = mysql_fetch_row($item_query);
 
			echo"You got an ".$item[0]."!";
 
		}
if( $rand_num <= $chance )
		{
			$item_id = mysql_query("SELECT `item` FROM `creatures` WHERE id = '$id_creature'");
			$foo =& creature_stats( item, $id_creature );
 
			$item_query = mysql_query("SELECT * FROM items WHERE id = '$foo'") or ('$item_query');
 
			$item = mysql_fetch_row($item_query);
 
			echo"You got an ".$item[0]."!";
 
		}

•

Join Date: May 2007

Posts: 68

Reputation: johny_d is an unknown quantity at this point

Solved Threads: 4

johny_d

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Junior Poster in Training

Re: Need Help with converting an ID to a Name

Oct 8th, 2007

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$item = mysql_fetch_array($item_query);

means $item is an array an you cannot put an array in an echo statement; this is why you get nothing.
You need to use $item['name_of_mysql_column'] in your "echo...."
You also have a sintax error: in

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$foo =& creature_stats( item, $id_creature );

since item is a variable, it needs a $ sign in front; this is a very good reason why you get a blank nothing

Last edited by johny_d; Oct 8th, 2007 at 5:39 pm.

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•

Join Date: Jun 2007

Posts: 1,227

Reputation: kkeith29 has a spectacular aura about

Solved Threads: 167

kkeith29

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Nearly a Posting Virtuoso

Re: Need Help with converting an ID to a Name

Oct 8th, 2007

sorry i overlooked that sintax error. wow i can't believe i missed that. also, before posting the code i did i should of asked the question; "Will the query ever return more than one result?". i figured there would only be one result and my way would work perfectly, but if more than one is returned you would need to use johny_d's way.