Printable View
Hi, maybe someone can help me with this...
I have
include('db_conn.php');
$queryget = mysql_query("SELECT * FROM reg_users ORDER BY `time` LIMIT 10") or die("Error with query");
while ($row = mysql_fetch_array($queryget))
{
$date = $row['date'];
$time = $row['time'];
$datet = date("Y-m-d");
$timet = date("H:i:s");
time's are saved as you can see: H:i:s (23:10:55)
so, i need to convert $time = $row['time']; and $timet = date("H:i:s"); to seconds... is there an option? THANKS
You can do this two ways (I suggest the second)
#1
(To get seconds:)
#2 (highly suggested)PHP Code:$seconds = explode(':', $timet)[2];
Instead of saving times as H:i:s, I would save times($row['time']) using thetime()function. Then when getting the time as H:i:s you would do:
Feel free to ask questions.PHP Code:$time = date("H:i:s", $row['time']);
echo $time; // 23:10:55
$seconds = date("s", $row['time']);
echo $seconds; //55
what does it means [2] ?? because of that i got an error and i cant open my page
Sorry. :/
PHP Code:$seconds = explode(':', $timet);
$seconds = $seconds[2];
it works, but takes only seconds. I need all time convert to seconds like 2H 5m 32s = 7532 seconds :)
what i have to write?
Haha, sorry:
It is not tested!Code:$total_seconds = 0;
$seconds = explode(':', $timet);
$total_seconds += $seconds[0] * (60 * 60);
$total_seconds += $seconds[1] * 60;
$total_seconds += $seconds[2];
echo $total_seconds;
Yea, it's working :)
Hi,
I faced problem of this function, but don't know where :)
I use this: [also I update their mysql time in every move]
and no users are online ALWAYS.PHP Code:...
function timeDiff($firstTime,$lastTime) {
$firstTime=strtotime($firstTime);
$lastTime=strtotime($lastTime);
$timeDiff=$lastTime-$firstTime;
return $timeDiff;
echo "$timeDiff";
}
while ($row = mysql_fetch_array($queryget))
{
$nick = $row['nick'];
$time = $row['time'];
$timet = date("H:i:s");
$id = $row['id'];
if (timeDiff($time,$timet) <= 300) {
if (timeDiff($time,$timet) >= -1) {
$var = '1';
echo "
$nick user is online
";
}}}
if ($var < 1) {echo "noone is online";}
Thanks :)
First of all
Onc you return something, the whole function is stopped.PHP Code:return $timeDiff;
echo "$timeDiff";
Second of all
Whats the point of that, just do:PHP Code:if (timeDiff($time,$timet) <= 300) {
if (timeDiff($time,$timet) >= -1) {
Next,PHP Code:if ((timeDiff($time,$timet) <= 300) AND timeDiff($time,$timet) >= -1) {
Even though PHP works with strings as numbers, you may want to do:PHP Code:$var = '1';
Instead.PHP Code:$var = 1;
And finally - what problem are you facing?
I don't know it just worked, but now something is wrong. What I was trying to create was online system. I have two times and two dates: $time & $date - current time and date; $timet & $datet - time and date when user made move in past or maybe now.
I need to count seconds of $time, $date and $timet, $dated to see which is higher.
if xtimetdated - xtimedate >= -300 it will show online.
time formats:
$datet = date("Y-m-d");
$timet = date("H:i:s");
same as $date and $time
I can't help until I know what is wrong. Can I see an example to your page, also can you please tell me whats supposed to happen from a users point of view.
Thanks! ~Nile
Here in the end of page... Every USER will see online nicks of others in green color. When theres no user online it shows black message
So - whats the problem? You don't want the black message?
The problem is in all system... I give a good time everytime, but somehow its not working. My problem is, that when i log into it shows that I am online and after 6 minutes its not. Somehow, time is counting in bad way...
Sorry - I didn't completely understand. 6 minutes after you log in - it shows whos online?
yes, 6 minutes it shows that im online. also I can see other online users.
can you understand my script which i posted? what it does...
Well, I can't help you, I'm very sorry. Try making a new thread - because this thread is old, and users don't want to post in it [also, the question your asking doesn't have anything to do with the original thread]
Sorry and good luck!
Ok, let's go back few steps... :)
I have
$date = date("Y-m-d");
$time = date("H:i:s");
How I can count(convert) date and time into seconds?
Ok, let's go back few steps... :)
I have
$date = date("Y-m-d");
$time = date("H:i:s");
How I can count(convert) date and time into seconds?
secondsConvert will return (hours*60*60)+(minutes*60)+seconds.PHP Code:<?php
function secondsConvert($time){
$seconds = explode(':', $time);
$total_seconds = 0;
$total_seconds += $seconds[0] * (60 * 60);
$total_seconds += $seconds[1] * 60;
$total_seconds += $seconds[2];
return $total_seconds;
}
$time = date("H:i:s");
echo secondsConvert($time)." seconds";
?>
how to do with date?
Well - for days, that would just be day * 24 * 60 * 60, but for months and years, it'd be hard because there aren't always four weeks in a month...
how to create difference then?
can I use $time - $timet = 300seconds ?
OR
if (($time - $timet) >= '00:05:00') then {echo "he is online";}
How to do something like that?
What do you mean? Can't you just insert names of the people online, and when the user logs out, there not online?
But sometimes people dont push log out and name would be still there
Hmm.. So every time the user goes somewhere, have a field for each user called "Minute" and insert the current minute. Then, when every user refreshes the page, it checks each users minutes, and if its more then 5 minutes idle, delete them from the list. (also, make sure 61 = 0, 62 = 1...)
Will this work even then when they close their browsers??
Yes - 5 minutes after they close the browser, they shold be off the list.
One more thing, if noone refreshes that page until tomorrow same time? page will not delete it...
Can someone tell me how the online system works on other sites??
Well. :rolleyes: If no one is using/on the site - no one will be looking at the whos online page, so there's not even a point of removing the user when no ones online.
eg: two(one, three) users closes their IE windows likely at the same time 10:30.
Then noone goes online until next day same time...
If i will go online, I will see them online... (but they are logged off)
5 minutes doesnt know what day is today... and it will be shown everyday same time. Offcourse, its quite impossible that noone will come online in one day, but still - bug :D ;D
Ok, so you would insert:
14 - 9:50
In the database, 14 being the date (thurs, jan. 14th), and 9:50 being the time. If the current date doesnt equal 14, there not online.