フレームワークcodeigniterを使用した検索機能、PHP

Question

私の問題は私が入力するたびにfnameとmname私はデータベースからエラーを受け取りました。検索フィールドにfnameとmnameの両方を検索したいだけです。

私searchコントローラ機能

public function search() 
    { 
    $li = $this->session->userdata('logged_in'); 
    $id = $this->session->userdata('idnumber'); 
    if($li == TRUE) 
    { 
     $this->load->model('users_model'); 
     $this->load->helper('smiley'); 
     $this->load->library('table'); 
     $image_array = get_clickable_smileys('http://localhost/efg/images/smileys', 'status'); 
     $col_array = $this->table->make_columns($image_array, 8); 
     $image_array2 = get_clickable_smileys('http://localhost/efg/images/smileys', 'status'); 
     $col_array2 = $this->table->make_columns($image_array2, 8); 
     $this->data['smiley_table1'] = $this->table->generate($col_array); 
     $this->data['smiley_tables'] = $this->table->generate($col_array2);   
     if($this->input->post()) 
     { 
     $search = $this->input->post('search'); 
     $memb = $this->users_model->search($search); 
     $usersearch = $this->users_model->search($search); 
     $grpsearch = $this->users_model->searchgrp($search); 
     redirect ('home/profile/'.$memb); 
     } 
    } 
} 

マイusers_modelモデル

public function search($search) 
{ 
    $this->db->select('*'); 
    $this->db->from('users'); 
    $this->db->like('username',$search); 
    $this->db->or_like('fname',$search); 
    $this->db->or_like('lname',$search); 
    $this->db->or_like('mname',$search); 
    $query = $this->db->get(); 
    foreach ($query->result_array() as $row) 
    { 
     $memb = $row['idnumber']; 
    } 
    $error = 'There is no record for the one you searched. Please go Back.'; 
    $query1 = $this->db->query("SELECT * FROM users WHERE idnumber ='$memb'"); 
    $hehe = $query1->result_array(); 
    if($hehe==NULL) 
    { 
    echo $error; exit; 
    } 
    else 
    { 
    return $memb; 
    } 
} 

Answer 1

必ず状態で使用してvariblesを確認し、条件の前に宣言されています。 それはyoullのは、クエリで使用する変数が設定されていることを確認し、出力

 public function search($search) 
     { 
      $this->db->select('*'); 
      $this->db->from('users'); 
      $this->db->like('username',$search); 
      $this->db->or_like('fname',$search); 
      $this->db->or_like('lname',$search); 
      $this->db->or_like('mname',$search); 
      $query = $this->db->get(); 
      $memb = null; 
      foreach ($query->result_array() as $row) 
      { 
       $memb = $row['idnumber']; 
      } 
      $error = 'There is no record for the one you searched. Please go Back.'; 
      $query1 = $this->db->query("SELECT * FROM users WHERE idnumber =$memb"); 
      $hehe = $query1->result_array(); 
      if($hehe==NULL) 
      { 
      return $error; 
      } 
      else 
      { 
      return $memb; 
      } 
     } 

Answer 2

どのようになるかの条件を渡すdoesntの場合。クエリ実行前のトラップ また、できる場合は、クエリビルダ（アクティブレコード）を使用してください。

public function search($search) 
{ 
    $this->db->select('*'); 
    $this->db->from('users'); 
    $this->db->like('username',$search); 
    $this->db->or_like('fname',$search); 
    $this->db->or_like('lname',$search); 
    $this->db->or_like('mname',$search); 
    $query = $this->db->get(); 

    $memb = ''; 
    if ($query->result_array() > 0) { 
     foreach ($query->result_array() as $row) { 
     $memb = $row['idnumber'];   
     } 
     if ($memb) { 

     $this->db->select("*"); 
     $this->db->where("idnumber", $memb); 
     $query1 = $this->db->get("users"); 

     //$query1 = $this->db->query("SELECT * FROM users WHERE idnumber ='$memb'"); 

     $hehe = $query1->result_array(); 
     return ($hehe) ? $memb : false; 
     } 
     return false; 
    } 

    $error = 'There is no record for the one you searched. Please go Back.'; 
    return $error; 
}