PHP返回所有行而不是请求的行
我有一个查询,它从表中返回所有行,而不是我指定的字段。你能看出这里面有什么错误吗?我用的是codeigniter 提前谢谢PHP返回所有行而不是请求的行,php,codeigniter,Php,Codeigniter,我有一个查询,它从表中返回所有行,而不是我指定的字段。你能看出这里面有什么错误吗?我用的是codeigniter 提前谢谢 unset($conditions); $conditions['conditions'] = array("accountid"=>$this->sessionInfo['database_account_id'], "DATE_FORMAT(salestart,'%Y-%m-%d')"=>$today,
unset($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);
我的模型是:
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();
}
改变这个
$this->Database->select('*');
对此
$this->Database->select($fields);
不知道-1是什么意思如果我遗漏了什么,请告诉我,谢谢!
$this
指的是什么?什么是Database\u Model
?我已经包括了我的Database\u Model…使用$this->Database->select($fields,true)代码>谢谢!现在工作。