php - CodeIgniter 中带有 if 语句的两个插入查询

Question

我试图通过检查用户类型在两个表中添加信息。如果用户是普通用户，则必须在第一个表中输入记录，如果用户类型是医生，则必须在两个表中输入记录。我正在使用 CodeIgniter 3.6

public function model_signup()
{
$nameVar                = $this->input->post("signup_name");
$emailVar           = $this->input->post("signup_email");
$phoneVar           = $this->input->post("signup_phone");
$passwordVar            = $this->input->post("signup_password");
$ifDoctorVar            = $this->input->post("signup_ifdoctor");
$pmdcVar                = $this->input->post("signup_pmdc");

$signu_query            = $this->db->query("SELECT * FROM `doc_users` WHERE `user_email`='".$emailVar."'");
if($signu_query->num_rows() >0){
return false;
}
else
{

   $this->db->query("INSERT INTO `doc_users`
   (`user_nicename`,
   `user_login`,
   `user_email`,
   `display_name`,
   `user_phone`,
   `user_pass`,
   `user_type`,
   `user_registered`)
   VALUES ('$nameVar',
   '$emailVar',
   '$emailVar',
   '$nameVar',
   '$phoneVar',
   '$passwordVar',
   '$ifDoctorVar',
   NOW())");

    $user_id = $this->db->insert_id();

   if($ifDoctorVar=='Doctor') {    // checking if user is doctor or not

   $this->db->query("INSERT INTO `doc_doc_details`
   (`ID`,
   `pmdc_id`,
   `email`,
   `phone)
   VALUES ('$user_id',
   '$pmdcVar',
   '$emailVar',
   '$phoneVar')");
  return true;
   }

  }
 }

Answer 1

public function model_signup()
{
$nameVar                = $this->input->post("signup_name");
$emailVar           = $this->input->post("signup_email");
$phoneVar           = $this->input->post("signup_phone");
$passwordVar            = $this->input->post("signup_password");
$ifDoctorVar            = $this->input->post("signup_ifdoctor");
$pmdcVar                = $this->input->post("signup_pmdc");

$signu_query            = $this->db->query("SELECT * FROM 
`doc_users`     WHERE `user_email`='".$emailVar."'");
if($signu_query->num_rows() >0){
return false;
}
else
{

$this->db->query("INSERT INTO `doc_users`
(`user_nicename`,
`user_login`,
`user_email`,
`display_name`,
`user_phone`,
`user_pass`,
`user_type`,
`user_registered`)
VALUES ('$nameVar',
'$emailVar',
'$emailVar',
'$nameVar',
'$phoneVar',
'$passwordVar',
'$ifDoctorVar',
 NOW())");

 $user_id = $this->db->insert_id();

 if($ifDoctorVar=='Doctor') {    // checking if user is doctor or not

 $this->db->query("INSERT INTO `doc_doc_details`
 (`ID`,
  `pmdc_id`,
  `email`,
  `phone)
  VALUES ('$user_id',
  '$pmdcVar',
  '$emailVar',
  '$phoneVar')");
    // return true remove this
  }

 }
}