我想将数组中的值保存到变量。进行if条件检查。这是我在models文件夹中的代码。我得到一个errormysqli_fetch_array()期望参数1是mysqli_result,给定对象。但是$ query返回一个数组值。
public function this_is_try(){
$this->db->select('*');
$this->db->from('partnerprofile');
$this->db->where('User_Id','20');
$query = $this->db->get();
$query->result_array();
$row = mysqli_fetch_array($query);
$user_id = $row['User_Id'];
$agefrom = $row['AgeFrom'];
print $user_id;
exit;
}
答案 0 :(得分:1)
而不是传递整个$ results(结果对象)。将result_object存储到变量并传递它,即$ result = $ results [0];
public function this_is_try(){
$this->db->select('*');
$this->db->from('partnerprofile');
$this->db->where('User_Id','20');
$query = $this->db->get();
$results = $query->result();
$result = $results[0];
$user_id=$result->user_id;
$agefrom = $result->AgeFrom;
print $user_id;exit;
}
答案 1 :(得分:0)
试试这个:
public function this_is_try(){
$row = $this->db->select('*')
->from('partnerprofile');
->where('User_Id','20');
->get()->row();
$user_id = $row->User_Id;
$agefrom = $row->AgeFrom;
print $user_id;exit;
}