Restricting multiple logins
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Restricting multiple logins
We are developing an application using PHP, MySQL. We would want to restrict multiple users logging in using the same username and password, at the same time. i.e., if I have logged in using the username and pwd: admin, and at the same time someone else uses that username and pwd, it should restrict. One of the ways we thought was, to set a flag as logged in when the first users logs in, so whenever any one else tries to log, based on the flag setting it will restrict. But now, if the first user closes the browser without logging out, then the flag remains as logged in the db and no one else can log in from there. How can we take care of this? Any help on this would be appreciated. Thanks
I guess a way around that would be to check the previous id/key and give it priority over any subsequent ones generated.
there might be another/better way to do this, but I don't know of any as most web apps stateless.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("e;SELECT id FROM table WHERE username = '"e;.$_POSTї'username']."e;' AND timelastactive >= '"e;.time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;.$_POSTї'username']."e; is already loggn the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("e;SELECT id FROM table WHERE username = '".$_POST['username']."e;' AND timelastactive >= '"e;.time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) &amto update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '"e;.$_POSTї'username']."e;' AND timelastactive >= '"e;.time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User &e load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$a field in your database that records the time a user was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).ere's not a way to tell what the user is actually doing (client side).on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '"e;.time()-3he time a user was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '"e;.time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).ur database that records the time a user was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).utes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;.$_POSTї'username']."e; is already logge likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).a field in your database that records the time a user was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side). records the time a user was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;tivity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;.$_POST['username']." is already logged into the system.");
}When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, d FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell whaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;.$_POSTї'username']."e; is already logged into the system.");
}
a field in your database that records the time a user was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '"e;.time()-300."e;'"e;);
// tser was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("e;SELECT id FROM table WHERE username = '"e;.$_POSTї'username']."e;' AND timelastactive >= '"e;.time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;.$_POSTї'username']."e; is already logged into the system."e;);
}When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}[/php:1:'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POSTї'username']."e;' AND timelastactive >= '"e;.time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).e if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '"e;.time()-300.& that records the time a user was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User "e;.$_POSTї'username']."e; is already logged into the system."e;);
}// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) &gn determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("e;SELECT id FROM table WHERE username = '"e;.$_POSTї'username']."e;' AND timelastactive >= '"e;.time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;.$_POSTї'username']."e; is already logged into the sysoad, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}[mine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("e;SELECT id FROM table WHERE username = '"e;.$_POSTї'username']."e;' AND timelastactive >= '"e;.time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;.$_POSTї'username']."e; is already lte. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("e;SELECT id FROM table WHERE username = '"e;.$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).E username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'"e;);
// this example is y to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)st active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '"e;.time()-300."etime a user was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("e;SELECT id FROM table WHERE username = '"e;.$_POSTї'username']."e;' AND timelastactive >= '"e;.time()-300."e;'"e;);
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;.$_POSTї'usrhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User ".$_POST['username']." is already logged i are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("e;User "e;.$_POSTї'username']."e; is already logged into the system."e;);
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).st active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database like
Code: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}[/php:1:ba7f00user was last active on the site. Perhaps on every page load, run a query to update this field.
Then determine if they are 'logged on' as say.. 15 minutes of time with no activity?
When you go to log on, query the database likeCode: Select all
$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}$query = mysql_query("SELECT id FROM table WHERE username = '".$_POST['username']."' AND timelastactive >= '".time()-300."'");
// this example is 5 minutes
// Check to see if the query was met
if(mysql_num_rows($query) > 0)
{
die("User ".$_POST['username']." is already logged into the system.");
}
This is about the only way to do it since PHP is server side, and there's not a way to tell what the user is actually doing (client side).
* This script makes sure a user can logged in only once at the same time.
*
* If the user requests a page, isValid should be called.
* This function compares the uid and cnt in the session with the uid and cnt in the database.
*
* If a user logs in, validate should be called.
* This function tests if the uid and pwd are valid, and if they are, the cnt is incremented.
*
* This way:
* If a user tries to login with a valid uid and pwd, he get's access. (No messing with session_timeout etc)
* If a user has logged in again, all the older sessions will have an invalid cnt.
-
fastfingertips
- Forum Contributor
- Posts: 242
- Joined: Sun Dec 28, 2003 1:40 am
- Contact:
In this case you will have to check: time stamp and ip of the user. If the session is expired you should allow a user to login otherwise the user should wait untill time stamp expires. Ofc you will have some problems with the dial up users (if they reconnect they probably will have a different IP) but this cases are just a few.