$ account_types = $ this-> members_model-> getAllAccountTypes();
$result_set = array();
foreach($account_types as $account_type){
$result_set[$account_type->db_name] = array(
'subscribed' => $this->members_model->getAllSubscribedMembers($account_type->id),
'unsubscribed' => $this->members_model->getAllUnsubscribedMembers($account_type->id)
);
}
echo json_encode($ result_set);
返回;
我有这些代码行返回我试图在foreach上获取非对象的属性($ account_types为$ account_type)
答案 0 :(得分:0)
更改您的代码,它将解决您的问题: -
$result_set = array();
$account_types = json_decode(json_encode($account_types), true);
foreach($account_types as $account_type){
$result_set[$account_type->db_name] = array(
'subscribed' => $this->members_model->getAllSubscribedMembers($account_type->id),
'unsubscribed' => $this->members_model->getAllUnsubscribedMembers($account_type->id)
);
}
答案 1 :(得分:0)
始终在模型中处理您的查询结果:
这可能是查询返回null,因此您最好先测试结果:
if ($query->num_rows() > 0)
{
foreach ($query->result() as $row)
{
echo $row->title;
echo $row->name;
echo $row->body;
}
}
您还可以将字符串传递给result()
,它表示要为每个结果对象实例化的类。 More Information