Question

任何人都可以在此查询中看到明显的错误吗？

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());
}

由于

编辑，错误讯息......

警告：mysql_fetch_assoc（）：提供的参数不是有效的MySQL结果资源

Answer 1

ateam.team AS awayteam,

不

ateam,team AS awayteam,

查询不正确？

1 个答案: