Help with SQL error

[CagedApe]

I've written out the following, but it returns a syntax error near the bolded portion. I don't see anything wrong with the syntax, but I'm sure theres something I missed. The script is supposed to scan the table to see if you have attacked your current target player 3 times in the last hour.

 

 

$q = mysql_query("SELECT COUNT(*) AS count FROM attack WHERE attacker_id = $AttackerId AND

defender_id = $AttackedID AND

attacktime BETWEEN DATE_SUB(NOW(), INTERVAL 1 HOUR) AND NOW()") or die(mysql_error());

$row = mysql_fetch_array($q) or die(mysql_error());

 

if($row['count'] == 3){

die("You have already attacked this person 3 times this hour.");

[Pikachu2000]

Without the ability to echo the query, you can't really debug it. Separate the query from the query execution, and echo it along with mysql_error().

 

$q = "SELECT COUNT(*) AS count FROM attack WHERE attacker_id = $AttackerId AND defender_id = $AttackedID AND attacktime BETWEEN DATE_SUB(NOW(), INTERVAL 1 HOUR) AND NOW()"
mysql_query($q) or die('<br />Query string: ' . $q . '<br />Produced error: ' . mysql_error() . '<br />');

[CagedApe]

Thanks. That pointed out my error and now the select command isn't giving an error. However, now its telling me mysql_fetch_array(): supplied argument is not a valid MySQL result resource. Any ideas?

 

 

$q = "SELECT COUNT(*) AS count FROM attack WHERE attacker_id = $AttackerId AND defender_id = $AttackedId AND attacktime BETWEEN DATE_SUB(NOW(), INTERVAL 1 HOUR) AND NOW()";
mysql_query($q) or die('<br />Query string: ' . $q . '<br />Produced error: ' . mysql_error() . '<br />');
$row = mysql_fetch_array($q) or die(mysql_error());

if($row['count'] == 3){
    die("You have already attacked this person 3 times this hour.");
} 

[jcbones]

$result = mysql_query($q) or die('<br />Query string: ' . $q . '<br />Produced error: ' . mysql_error() . '<br />');
$row = mysql_fetch_array($result) or die(mysql_error());

if($row['count'] == 3){
    die("You have already attacked this person 3 times this hour.");
} 

[Pikachu2000]

Changed the structure to supply mysql_fetch_assoc() with a valid result . . .

$q = "SELECT COUNT(*) AS count FROM attack WHERE attacker_id = $AttackerId AND defender_id = $AttackedId AND attacktime BETWEEN DATE_SUB(NOW(), INTERVAL 1 HOUR) AND NOW()";

$result = mysql_query($q) or die('<br />Query string: ' . $q . '<br />Produced error: ' . mysql_error() . '<br />');

$row = mysql_fetch_assoc($result) or die(mysql_error());

if($row['count'] == 3){
    die("You have already attacked this person 3 times this hour.");
}

[CagedApe]

Thank you Pikachu! That works perfectly.