如何編寫MySQL查詢像笨

Question

聯接如何笨

$query=$this->db->query("select TypeID,verification_type from table_verify where TypeID not in(select verificationId from table_invistigate where Id=$id)"); 
      $result = $query->result(); 
     return $result; 

Answer 1

像下面寫像一個聯接條件查詢該this.first得到驗證ID的數組，那麼在$this->db->where_not_in()使用這個數組。

$this->db->select('verificationId'); 
$this->db->where('Id',$id); 
$verificationIds = $this->db->get('table_invistigate')->result_array(); //array of verificationIds 

foreach($verificationIds as $vid){ 
$ids[] = $vid['verificationId']; 
} 


$this->db->select('TypeID,verification_type'); 
$this->db->where_not_in('TypeID',$ids); 
$result = $this->db->get('table_verify')->result_array(); 
print_r($result); 

Answer 2

$this->db->select('table_verify.*,table_invistigate.verificationId'); 
    $this->db->from('table_verify'); 
    $this->db->join('table_invistigate', 
    'table_verify.verificationId=table_invistigate.verificationId '); 
    $this->db->where('table_verify.id',$id); 
    $query = $this->db->get(); 
    return $query->result();