这是的三个表
1. table `p_transactions` has following fields
(`txn_id`, `txn_uid`, `txn_bid_no`, `txn_date`, `txn_desc`, `txn_amt`, `txn_fee`, `txn_mode`, `txn_status`, `txn_mdate`)
2 . table `p_game_results` has following fields.
(`id`, `game_id`, `game_combo`, `game_combo_hr`, `cdate`, `mdate`)
3. Table `p_game_room_results` has following fields
(`id`, `game_id`, `room_id`, `txn`, `round`, `result`, `score`, `cdate`, `mdate`)
我想加入他们使用他们的txn id作为一个公共字段。
这是我试过的东西。但不确定,我肯定是错的。
$sql = "SELECT * FROM " . $prefix . "_user_game_results".$prefix."_game_room_results".$prefix."_transactions WHERE". $prefix . "_user_game_result.uid"='$prefix."_game_room_results.id"'. and .$prefix."_transactions.txn_uid"='$prefix."_game_room_results.uid"'"";
$result = $this->sql_fetchrowset($this->sql_query($sql));
谢谢。
这应该做到:
select * from p_transactions pt
inner join p_game_room_results pgrr on pgrr.txn=pt.txn_id
inner join p_game_results pgr on pgr.game_id=pgrr.game_id
按公用字段联接所有表。CCD_ 1和CCD_。
select *
from p_transactions as p_t INNER JOIN p_game_results as p_g_r
ON p_t.txn_id = p_g_r.id
INNER JOIN
p_game_room_results as p_g_r_r
ON p_g_r_r.game_id = p_g_r.game_id;
以下别名用于INNER JOIN
p_transactions as p_t
p_game_results as p_g_r
p_game_room_results as p_g_r_r
我建议您阅读这些有用的SQL Joins教程
http://www.codinghorror.com/blog/2007/10/a-visual-explanation-of-sql-joins.html
http://en.wikipedia.org/wiki/Join_%28SQL%29