任何人都可以看到此查询中的明显错误吗?
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());
}
谢谢
编辑,错误消息...
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource
最佳答案
ateam.team AS awayteam,
不是
ateam,team AS awayteam,
关于mysql - 查询不正确?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4642730/