Question

我在数据库中有3个表：

团队

id

名称

匹配

id（int）

team_home_id

team_away_id

目标

id

match_id

team_id

时间

我需要在控制器中获得目标的情况下显示团队的名称。

我知道我应该做联接表。

我有此代码：

    public function get_goals() {
        $this->db->select('goals.*');
        $this->db->from('goals');
        $this->db->join('matches', 'matches.id = goals.match_id');
        $q = $this->db->get();

        return $q->result();
    }

我不知道接下来要怎么做。

我在视图显示名称中需要：

$ goals-> team_home_name 和 $ goals-> team_away_name

Answer 1

您可以尝试以下操作：-

public function get_goals() {
    $this->db->select('g.* , t.name as team_home_name , t.name as team_away_name);
    $this->db->from('goals as g');
    $this->db->join('matches as m', 'm.id = g.match_id');
    $this->db->join('teams as t', 't.id = g.team_id');
    $q = $this->db->get();
    return $q->result();
}

Answer 2

要获得$goals->team_home_name和$goals->team_away_name的结果，请使用如下别名：

public function get_goals() {
    $this->db->select('goals.*, home_team.name team_home_name, away_team.name team_away_name');
    $this->db->from('goals');
    $this->db->join('matches', 'matches.id = goals.match_id');
    $this->db->join('teams home_team', 'home_team.id = matches.team_home_id');
    $this->db->join('teams away_team', 'away_team.id = matches.team_away_id');
    $q = $this->db->get();

    return $q->result();
}

Answer 3

U可以使用别名

从“团队”表中获取team_away_name和team_home_name

public function get_goals() {
        $this->db->select('goals.*, team.team_home_name, team.team_away_name');
        $this->db->from('goals');
        $this->db->join('matches', 'matches.id = goals.match_id');
        $this->db->join('teams as team', 'team.id = goals.team_id');
        $q = $this->db->get();

        return $q->result();
    }

如何使用别名联接多个表

3 个答案: