小编And*_*rew的帖子

我在我的查询上使用ON DUPLICATE UPDATE,一些结果没有存储它.我尝试了所有可能的方式,但那些仍然保持不变.这是数据库图片.

那些NULL,是没有成功存储的行; 结果应为1而不是NULL.

    if($remark){
$query3 = "INSERT INTO `audit_section_remarkrecord` SET remark = '$remark', form_details_subquestion_id = '$form_details_subquestion_id', form_details_section_id = '$form_details_section_id', audit_section_no = '$audit_no' ON DUPLICATE KEY UPDATE
form_details_section_id = '$form_details_section_id' , remark = '$remark'";
$result3 = $db->query($query3); 

$query4 = "UPDATE `remarkrecord_update_details` SET form_details_section_id = '$form_details_section_id', userlog = '$user_staff', ipaddress = '$ip' WHERE form_details_subquestion_id = '$form_details_subquestion_id' AND audit_section_no = '$audit_no' ";
$result4 = $db->query($query4); 
}else{
}

}

表结构

目前我想在用户登录系统时从两个表中获取两个不同的数据.权利表是"user","user_staff"和"user_group".但是当用户输入其用户名和密码并提交它时,提示"致命错误:在第26行的C:\ xampp\htdocs\auditsystem\index.php中的非对象上调用成员函数fetch_array()"

以下是代码:

if($username!= "" && $password != "")
{
//INNER JOIN user_group_module_role ON user.user_group_module_role_id = user_group_module_role.id
    $result =  $db->query("SELECT * FROM user 
                            INNER JOIN user_staff ON user.user_staff_id = user_staff.id
                            WHERE username = '$username' AND password = '$password'"); 



    if($result->num_rows == 1)
    {
        $validate = $result->fetch_assoc();

        $query1 = "SELECT * FROM usergroup WHERE id = $validate[user_group_id]";
        $result1 = $db->query($query1);
        $row1 = $result1->fetch_array();
        //change here for the authority 
        $_SESSION['user_staff'] = $validate['displayname'];
        $_SESSION['usergroup'] = $row1['user_group_type'];
        echo "<script language='javascript'>window.location='panel.php'</script>"; 
    }
    else
    {
        echo …