查询不正确？

任何人都可以在这个查询中看到一个明显的错误？查询不正确？

function getFixtureDetails($league, $date, $status) 
{ 
    global $database; 
    $q = "SELECT g.id, g.home_user, g.home_user2, g.away_user, g.away_user2, g.home_score, g.away_score, hteam.team AS hometeam, ateam,team AS awayteam, 
      huser.username AS home_username, huser2.username AS home_username2, auser.username AS away_username, auser2.username AS away_username2 
      FROM ".TBL_FOOT_GAMES." g 
      INNER JOIN ".TBL_FOOT_TEAMS." hteam ON hteam.id = g.home_team 
      INNER JOIN ".TBL_FOOT_TEAMS." ateam ON ateam.id = g.away_team 
      INNER JOIN ".TBL_USERS." huser ON huser.id = g.home_user 
      LEFT JOIN ".TBL_USERS." huser2 ON huser2.id = g.home_user2 
      INNER JOIN ".TBL_USERS." auser ON auser.id = g.away_user 
      LEFT JOIN ".TBL_USERS." auser2 ON auser2.id = g.away_user2 
      WHERE g.fixture_date = '$date' AND g.leagueid = '$league' AND (g.type = '2' OR g.type = '12' OR g.type = '22' OR g.type = '32') AND g.status = '$status' 
      ORDER BY g.fixture_date"; 
    return mysql_query($q, $database->myConnection()); 
}

感谢

编辑，错误信息...

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource

来源

2011-01-10 sark9012

为什么不包括你得到错误信息？ – ChristopheD 2011-01-10 00:14:36

添加了错误消息 – sark9012 2011-01-10 00:16:17

ateam.team AS awayteam,

不

ateam,team AS awayteam,

来源

2011-01-10 00:20:19

查询不正确？

回答

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