Question

public function user_data_vi($email)
{
    $this->db->select('*');
    $this->db->from('client');
    $this->db->join('payments', "payments.email = $email");

    $q= $this->db->get();

    return $q->result(); 
}

上面的代码是我的模型和方法user_data_vi我在两个表client字段中有两个表payments而另一个email现在很常见我想要从数据库中检索数据第一个表具有用户信息，payment table具有付款信息，使用电子邮件字段获取两个表数据信息。

Answer 1

您需要在此处添加where条件，请在下面找到修改后的代码

public function user_data_vi($email)
{
        $this->db->select('*');
        $this->db->from('client');
        $this->db->join('payments', "payments.email = client.email");
        $this->db->where('client.email',$email);
        $q= $this->db->get();
        return $q->result(); 
}

Answer 2

public function user_data_vi($email)
{
    $this->db->select('*');
    $this->db->from('client');
    $this->db->where('client.email',$email);
    $this->db->join('payments','payments.email = client.email');
    $q= $this->db->get();
    return $q->result(); 

}

如何使用codeigniter中的join从两个表中获取值

2 个答案: