Question

我在PHPMYADMIN中有这些表，我想要的是获取每个团队使用ID的名称，我一直在尝试使用LEFT JOIN，但它只返回1个团队，否则它将返回null

此表匹配

        |   match_id    |   HomeTeam    |   AwayTeam    |
        +------------------------------------------------
        |       1       |       3       |       1       |

这是表团队

    |   team_id     |       team_name       |
    +----------------------------------------
    |       1       |       Manchester      |
    |       2       |       Barcelona       |
    |       3       |       Real Madrid     |

这就是我想要SQL返回的结果

    |match_id|HomeTeam|AwayTeam| Home_name   | Away_name |
    +-----------------------------------------------------
    |    1   |   3    |   1    | Real Madrid | Manchester|

我一直在尝试使用这个SQL语句

    SELECT `match.*`,`team_*` FROM `match`
    LEFT JOIN `team` ON `match.HomeTeam` = `team.match_id`
    LEFT JOIN `team` ON `match.AwayTeam` = `team.match_id`
    WHERE `match.match_id` = 1

任何解决方案？

目前我正在使用这个PHP代码，但我觉得效率不高

    $Query  = 'SELECT * FROM match WHERE match_id = 1'
    // excute Query //

    foreach($match_array as $match)
    {
        $Query = 'SELECT FROM team WHERE team_id = '.$match['HomeTeam'].''

        $home_team_name = somthing

        $Query = 'SELECT FROM team WHERE team_id = '.$match['AwayTeam'].''
                    $away_team_name = something 
    }

Answer 1

你必须使用表别名两次加入团队表。

SELECT match.*,team_home.*,team_away.* FROM match
  LEFT JOIN team team_home ON match.HomeTeam = team_home.team_id
  LEFT JOIN team team_away ON match.AwayTeam = team_away.team_id
  WHERE match.match_id = 1

Answer 2

您可以尝试使用别名以获取正确的信息：

SELECT `m`.`*`, `ht`.`name`, `at`.`name` FROM `match` m
LEFT JOIN `team` ht ON `m`.`HomeTeam` = `ht`.`match_id`
LEFT JOIN `team` at ON `m`.`AwayTeam` = `at`.`match_id`
WHERE `m`.`match_id` = 1

这可以给你你想要的东西

在2个字段上加入2个表

2 个答案: