PHP帶回所有行而不是請求

Question

我有一個查詢是從表中取回所有行，而不是我指定的字段。你能看到這個錯誤嗎？我正在使用codeigniter。

在此先感謝！

unset($conditions); 
    $conditions['conditions'] = array("accountid"=>$this->sessionInfo['database_account_id'], 
       "DATE_FORMAT(salestart,'%Y-%m-%d')"=>$today, 
       "shop"=>"london" 
       ); 
    $conditions['group_by'] = "item"; 
    $conditions['fields'] = "accountid, item, count(uniqueid) as totalitems, sum(options) as totaloptions, colour"; 
    $today_sales = $this->Database_Model->selectData("sales",$conditions); 

我的模式是：

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

更改此

$this->Database->select('*'); 

這個

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