A little problem, quick to fix.

[Bikkebakke]

So, sorry for asking such a simple question but I couldn't find any info on what causes this or what did I do wrong.

 

Heres my code:

$monsterID = rand(10,10);
$monster = mysql_query("SELECT name FROM monsters WHERE id = 'like {$mosnterID}'");

And:

echo "You encounter {$monster}";

 

Shows as:

You encounter Resource id #4

 

I want it to return the name of the monster by looking up the ID of the monster.

 

Is it my db table or is the mysql_query invalid or something?

 

Thanks in advance and sorry for bothering you.

[ram4nd]

Go look for some mysql php tutorials. You mysql_query returns array.  You can see in it with print_r

[Bikkebakke]

I've read trough a few good tutorials over the week I've been studying PHP/MySQL but didn't find help, that's why I'm asking for some. You know it's hard to find a solution to a problem if you don't know where the problem is. 

 

Thanks for the answer, even if it doesn't help me at all. (Rather than telling whats happening you could tell whats the issue here)

 

 

[Bikkebakke]

Also, in the code:

$monster = mysql_query("SELECT name FROM monsters WHERE id = 'like {$mosnterID}'");

 

{$mosnterID} is typoed, it is monsterID, that's not the problem.

[Bikkebakke]

Okay I figured it out myself:

 

$monsterID = rand(10,10);
$monsterquery = mysql_query("SELECT * FROM monsters WHERE id = $monsterID");
$monster = mysql_fetch_array($monsterquery);

and:

echo "You encounter {$monster['name']}";

[aeroswat]

You encounter resource ID #4... sounds scary

[ram4nd]

try something like

$monsterID = 10;
$monsterquery = mysql_query('SELECT * FROM monsters WHERE id = '.$monsterID);
$monster = mysql_fetch_array($monsterquery);

foreach($monster as $row)
echo row['id'];