


A brief discussion on PHP+MYSQL authentication method_PHP tutorial
近日在为学校制做校友录时,需要身份验证,在对比之后决定采用PHP+MYSQL进行身份验证。
之前也曾考虑过用cookies或session。但是用cookies,在用户离线再上线后,只要cookies不过
期,不用登录仍然可以保持在线,这对于网吧来说是个隐患。而且用户可以关闭cookies,这样身份验
证就不成功。也考虑过用session,session在浏览过程中不断的将访问信息加入到session中,如果用户
在网站内时间很长,浏览的页面很多,就用导致session越来越大,浏览速度降低,最后只有重新登录,
虽然这种情况不多见,但不是我们所希望的。
我在做这个身份验证时的想法是,在身份验证的同时,记录浏览信息。
用户ID在每个页面间传递,ID值是用MD5()函数加密得到的。验证函数是validate_id(),返回值为
(0,1),成功为“1”。
思路:
判断被传入的ID值是否为匿名登录ID(a684dd572b1887661782981659331eed),32位,如果是返回0,并且
将浏览信息加入数据库。如果否,则查询数据库,看数据库中的用户ID,用户IP与传入的ID,IP值是否相等
并且最近浏览时间距当前时间不到20分钟的记录。
判断得到的记录数,如果为0,则认为离线,并用匿名ID登录浏览信息,返回0。记录不为0时,将
用户ID,用户IP值,加入数据库,返回1。
创建数据库:
create table logging{
id int unsigned not null primary key auto_increment,
user_id char(32) not null,//用户ID
logging_ip varchar(20) not null,//记录用户IP地址
page_name varchar(30) not null,//浏览网页名
view_time timestamp not null,
student_id varchar(20)
);
创建函数:
/*-----begin function validate-id()---------------
验证用户是否登录
------------------------------------------------*/
function validate_id($link,$id,$ip,$page_name,$student_id=""){
if($id==a684dd572b1887661782981659331eed or $id=){
$query=insert into logging(user_id,logging_ip,page_name,student_id) values ("a684dd572b1887661782981659331eed",".$ip.",".$page_name.","anonym");;
$result=mysql_db_query("web",$query,$link);
return (0);
}
else {
$year=strftime("%Y");
$month=strftime("%m");
$day=strftime("%d");
$hour=strftime("%H");
$min=strftime("%M");
$sec=strftime("%S");
echo $time_string=$year.$month.$day.$hour.$min.$sec;
// echo ("
");
// echo "$year-$month-$day $hour-$min-$sec
";
file://---------begin ifs---------------------------
if(($min-=20)$min+=60;
if(($hour-=1)==-1){
$hour+=24;
if(($day-=1)==0){
switch ($month) {
case 12 :$days=30;break;
case 1 :$days=31;break;
case 2 :if(($year/4==0)and($year/100!=0)or($year/400==0)){
$days=29;}
else {$days=28;}
break;
case 3 :$days=31;break;
case 4 :$days=30;break;
case 5 :$days=31;break;
case 6 :$days=30;break;
case 7 :$days=31;break;
case 8 :$days=31;break;
case 9 :$days=30;break;
case 10 :$days=31;break;
case 11 :$days=30;break;
}
$day+=$days;
if(($month-=1)==0){
$month+=12;
$year-=1;
}
}
}
}
file://----------------------------------end ifs
setType($month,"integer");
if($monthsetType($month,"string");
$month=0.$month;}
setType($day,"integer");
if($daysetType($day,"string");
$day=0.$day;}
setType($hour,"integer");
if($hoursetType($min,"string");
$hour=0.$hour;}
setType($min,"integer");
if($minsetType($min,"string");
$min=0.$min;}
echo
.$time_string=$year.$month.$day.$hour.$min.$sec;
// echo "
$year-$month-$day $hour-$min-$sec
";
// echo ("
");
$query="select id from logging where user_id=$id and logging_ip=$ip and view_time>$time_string;";
$result=mysql_db_query("web",$query,$link);
$count=mysql_num_rows($result);
if($count==0){
// echo $query="insert into logging(user_id,logging_ip,page_name) values (a684dd572b1887661782981659331eed,$ip,$page_name);";
$result=mysql_db_query("web",$query,$link);
return (0);
}
else{
$query="insert into logging(user_id,logging_ip,page_name) values($id,$ip,$page_name)";
$result=mysql_db_query("web",$query,$link);
return (1);
}
}//end if
}
file://----------------------- -----end function validate-id---------
This verification method is very simple and does not take into account the user logging in again after logging in. You can add it yourself.
If you use cookies, you may use setcookies() to create a user ID, and then read it from the environment variable $HTTP_COOKIE or $HTTP_COOKIE_VARS
Take. It's all the same, but the user should be guaranteed to have no cookies.

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