要求の代わりにすべての行を返すPHP

Question

私が指定したフィールドではなく、テーブルからすべての行を戻すクエリがあります。これで何か間違いを見ることができますか？私はcodeigniterを使用しています。

ありがとうございます！

public function selectData($table,$condition=array()) { 
    if(isset($condition['fields'])){ 
     $fields = $condition['fields']; 
    } 
    else{ 
     $fields = "*"; 
    } 
    $this->Database->select('*'); 
    $this->Database->from($table); 
    if(isset($condition['conditions'])){ 
     $this->Database->where($condition['conditions']); 
    } 
    if(isset($condition['group_by'])){ 
     $this->Database->group_by($condition['group_by']); 
    } 
    if(isset($condition['order_by'])){ 
     $this->Database->order_by($condition['order_by']); 
    } 
    if(isset($condition['where_in'])){ 
     $where_in = $condition['where_in']; 


     foreach($where_in as $key =>$value){ 
      $this->Database->where_in($key,$value); 
     } 
    } 
    if(isset($condition['joins'])){ 
     $joins = $condition['joins']; 
     foreach($joins as $join){ 
      $this->Database->join($join['table'], $join['joinWith'],$join['type']); 
     } 
    } 
    $query = $this->Database->get(); 

    return $query->result_array(); 
} 

Answer 1

変更この

$this->Database->select($fields);