我有一个很大的问题,在网上阅读也许我已经找到了解决方案,但它的逻辑太难理解了,我想:)CREATE TABLE eventi (autore VARCHAR(100) NOT NULL,titolo VARCHAR(50) NOT NULL,data_evento VARCHAR(30) NOT NULL,
fine_evento VARCHAR(30
$attendingUsers = mysql_query("Select acceptedInvites from events where eventID = ".[$i]; //displays the userid number properly so I know this is working
$friendInfo = mysql_query