0

查询 1

$query = "SELECT  match_id, match_date, home_team, 
    home_score, away_score, away_team, rating

              FROM premier_league2 
              WHERE (home_team ='" .$home_team ."'
              OR    away_team ='" .$home_team ."')  
              AND   postponed !=1
              AND   league =1
              AND   match_date <'" . $current_date ."'
              AND   match_date >'" . $newseason ."'
              ORDER BY match_date DESC LIMIT 1";

结果

676 2013-05-19  Newcastle Utd   0   1   Arsenal -14

查询 2

SELECT home_per, draw_per, away_per FROM rating WHERE rating = -14;

结果

28.30 26.50 45.10

现在我正在尝试加入这些查询，例如

676 2013-05-19  Newcastle Utd   0   1   Arsenal -14  28.30 26.50 45.10

评分 (-14) 在这两个表中都很常见，寻求您的大力帮助。

4

2 回答 2

2

你可以通过加入来做到这一点。

SELECT 
    a.match_id, a.match_date, a.home_team, a.home_score, a.away_score, a.away_team, a.reting, b.home_per, b.draw_per, b.away_per 
FROM
    premier_league2 a
LEFT JOIN 
    rating b ON a.rating = b.rating
WHERE 
    (home_team = %HOME_TEAM%
    OR    away_team = %HOME_TEAM%)  
    AND   postponed !=1
    AND   league =1
    AND   match_date < %SURRENT_DATE%
    AND   match_date > %NEW_SEASON%
    ORDER BY match_date DESC LIMIT 1)

请不要mysql_*在新代码中使用函数。它们不再被维护并被正式弃用。看到红框了吗？改为了解准备好的语句，并使用PDO或MySQLi -本文将帮助您决定使用哪个。如果您选择 PDO，这里有一个很好的教程。

于 2013-06-26T19:06:38.497 回答
2 
SELECT
   pl.match_id, pl.match_date, pl.home_team, 
    pl.home_score, pl.away_score, pl.away_team, pl.rating,
   r.home_per, r.draw_per, r.away_per
FROM
   premier_league2 pl
 INNER JOIN rating as r ON pl.rating = r.rating
WHERE
(pl.home_team ='" .$home_team ."'
              OR    pl.away_team ='" .$home_team ."')  
              AND   pl.postponed !=1
              AND   pl.league =1
              AND   pl.match_date <'" . $current_date ."'
              AND   pl.match_date >'" . $newseason ."'
              ORDER BY pl.match_date DESC LIMIT 1";
于 2013-06-26T19:07:27.203 回答