具有多个表的复杂INNER JOIN查询的回显结果

您需要使用列别名并在提取调用中访问它们。而不是

event.*

$SQL = "SELECt sport.name AS sportname,country.name AS countryname, tournament_template.name AS templatename, tournament.name AS tournamentname, tournament_stage.name AS stagename, event.name AS eventname,event.someothercol AS someother FROM tournament_template ...etc......etc...";// Later...while($row=mysql_fetch_array($result)){  echo $row['eventname'];  echo "<br>";}