如何根据codeigniter中的手机号码从两个表中获取结果

Question

这里我有两个表，我想根据手机号码从这两个表中写一个选择结果。

第一个表

新员工

staff_id              firstName        Mobile          userType

  1                  Soupranjali       9986125566       Teacher
  2                  Sujata            8553880306       Teacher

第二张桌子

新学生

student_id           first_Name        fatherMobile        user_type

  1                  janarthan         8553880306         Student
  2                  Santanu           8277904354         Student

这里 8553880306 两个表都可用，所以我需要这样的输出

预期结果

    {
  "status": "Success",
  "Profile": [
    {
      "staff_id": "2",
      "firstName": "Sujata",
      "userType" : "Teacher"
    },
    {
      "student_id": "1",
      "firstName": "janarthan",
      "user_type" : "Student"
    },
  ]
}

如此尝试，但我无法得到答案，所以请任何人帮助我，

我的模特

 public function android_memberList($mobile)
            {
                $this->db->select('new_staff.staff_id, new_staff.firstName, new_staff.userType, new_student.student_id, new_student.first_Name, new_student.user_type');
                //$this->db->select('*');
                $this->db->from('new_staff');
                $this->db->join('new_student', 'new_student.fatherMobile  = new_staff.Mobile');
                $query = $this->db->get();

                # result_array is used to convert the data into an array
                $result = $query->result_array(); // or  $result = $query->result_array()[0]; 
                echo json_encode($query);
            }

更新答案

     [
  {
    "staff_id": "2",
    "firstName": "Sujata",
    "userType": "Teacher",
    "student_id": "1",
    "first_Name": "janarthan",
    "user_type": "Student"
  }
]

更新答案 2

    [
  [
    {
      "staff_id": "2",
      "firstName": "Sujata",
      "userType": "Teacher",
      "student_id": "1",
      "first_Name": "janarthan",
      "user_type": "Student"
    }
  ]
]

Answer 1

您需要检查两个表上移动列的数据类型。

它们必须具有相同的数据类型才能执行连接操作。

Answer 2

你可以试试这个

public function android_memberList($mobile)
{
    $this->db->select('new_staff.staff_id, new_staff.firstName, new_staff.userType, new_student.student_id, new_student.firstName, new_student.user_type');
    //$this->db->select('*');
    $this->db->from('new_staff');
    $this->db->join('new_student', 'new_student.fatherMobile  = new_staff.Mobile');
    $query = $this->db->get();

    # result_array is used to convert the data into an array
    $result = $query->result_array(); // or  $result = $query->result_array()[0]; 
    echo json_encode($result);
}

当您使用result_array 时，它将以零索引数组的形式出现。您可以选择它，也可以使用上述代码的注释将其删除