 PHP interesse |
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Ik heb:
$dbresxa = mysql_query("SELECT `id` FROM `[users]` WHERE UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(`online`) < 300 AND gameid={$_SESSION['gameid']}");
$onliner = mysql_num_rows($dbresxa);
$dbres = mysql_query("SELECT id FROM `[users]` WHERE gameid={$_SESSION['gameid']}");
$members = mysql_num_rows($dbres);
$member = mysql_fetch_object($dbres);
$select = mysql_query("SELECT login FROM `[users]` WHERE gameid={$_SESSION['gameid']} ORDER BY signup DESC LIMIT 0,1");
$object = mysql_fetch_object($select);
$dbresxa = mysql_query("SELECT `id` FROM `[users]` WHERE UNIX_TIMESTAMP(NOW())-UNIX_TIMESTAMP(`online`) < 300 AND gameid={$_SESSION['gameid']}"); $dbres = mysql_query("SELECT id FROM `[users]` WHERE gameid={$_SESSION['gameid']}"); $select = mysql_query("SELECT login FROM `[users]` WHERE gameid={$_SESSION['gameid']} ORDER BY signup DESC LIMIT 0,1");
Nou geeft mijn error log aan:
supplied argument is not a valid MySQL result resource
Betreft deze regels:
$onliner = mysql_num_rows($dbresxa);
en
$members = mysql_num_rows($dbres);
en
$member = mysql_fetch_object($dbres);
en tenslotte
$object = mysql_fetch_object($select);
verder werkt het script gewoon goed en zonder problemen. Wie kan mij vertellen hoe ik deze error toch kan oplossen?
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Wat is jouw favoriete javascript framework? (S: 18, R: 2)
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