这是我的第一个问题。我有一个复杂的SQL数据库,需要连接具有相同列名的不同表。
“赛事”是一场体育比赛。它包含链接到Tournament_stage表的Tournament_stageFK,后者包含链接到锦标赛表的TournamentFK,包含链接到Tournament_template表的Tournament_templateFK,包含包含SportPK的SportFK。
因此,为了找出比赛的来源,我需要进行内部联接,否则我将不得不打开数据库数百万次。唯一的方法就是这样做,但是我不知道如何显示结果。我对结果的回应很糟糕,如下所示:
$SQL = "SELECT sport.name, country.name, tournament_template.name, tournament.name, tournament_stage.name, event.* FROM tournament_template INNER JOIN sport ON tournament_template.sportFK = sport.id INNER JOIN tournament ON tournament.tournament_templateFK = tournament_template.id INNER JOIN tournament_stage ON tournament_stage.tournamentFK = tournament.id INNER JOIN event ON event.tournament_stageFK = tournament_stage.id INNER JOIN country ON tournament_stage.countryFK = country.id WHERE DATE(event.startdate) = CURRENT_DATE() ORDER BY sport.name ASC, country.name ASC, tournament_stage.name ASC, event.startdate ASC"; $result = mysql_query($SQL); while($get=mysql_fetch_array($result)) { echo $result['event.name']; echo "<br>"; }
您需要使用列别名并在提取调用中访问它们。而不是event.*,明确需要的列:
event.*
$SQL = "SELECT sport.name AS sportname, country.name AS countryname, tournament_template.name AS templatename, tournament.name AS tournamentname, tournament_stage.name AS stagename, /* Use explicit column names instead of event.* */ event.name AS eventname, event.someothercol AS someother FROM tournament_template ...etc... ...etc..."; // Later... while($row=mysql_fetch_array($result)) { echo $row['eventname']; echo "<br>"; }