PHP - Online In Last 24 Hour

Hi all,
I have a loop that is modifying a DateTime object. It goes through X number of times and adds 1 day to the date time object and stores the modified object in an array. I use ->add(new DateInterval('P1D')); For some reason it is losing an hour each time that I add a day. I start out at 19:00 and each iteration through the loop adds a day, but loses an hour.
2010-09-19 19:00:00
2010-09-20 18:00:00
2010-09-21 17:00:00

I can correct for this by doing ->add(new DateInterval('P1DT1H'));, but adding 25 hours instead of 24 hours doesn't make any sense to me. I am using PHP Version 5.3.1.

Any ideas?

Thanks,
W

Hello,

I have a very simple table... DAY (Y-m-d) / TIME (00:00:00) / POWER (INT)

I am using a Jquery datepicker to select the date on the page, and that POST to the PHP file.

I am trying to select values from Mysql to make 3 HighCharts Graphs using the selected day of the datepicker. I have started with the DAY graph PHP to get the values for each hour of a 24 hour day. I need to get the values for each hour... 23,12,15,35 etc... , and then I need for the 31 days of a month for the month graph, and 12 months of the year for the month graph all in the same way so the HighCharts can use the values to make the chart (3 php files, one for each graph)

Here is the PHP I have for the 1 day that should get the 24 individual hour data, but it does not seem to work...

<?php
$choice = (isset($_POST['choice'])) ? date("Y-m-d",strtotime($_POST['choice'])) : date("Y-m-d"); 
$con = mysql_connect("localhost","root","mackie1604");  
if (!$con)  {  
die('Could not connect: ' . mysql_error());  
}  
mysql_select_db("inverters", $con);  
$sql = "SELECT HOUR(time), COUNT(power) FROM feed WHERE time = '".$choice."' GROUP BY HOUR(time) ORDER BY HOUR(time) "; 
$res = mysql_query($sql) or die('sql='.$sql."\n".mysql_error()); 
$row = mysql_fetch_assoc($res); 
echo $row['choice'].'<br />'; 
?>

What have a written wrong in the sql query ???

Alan

So I am trying to check if the current time is one hour before a variable time:
Code: [Select]
$date_game=$dt->format('Y n j'.$pieces[2]);
echo $date_game;
echo date('Y n j H');

if (date('Y n j H') < $date_game)  {
echo "The time is before the stored time";
}

This displays
2012 1 21 17:30
2012 1 21 16
i.e $pieces[2] = 17:30. and $dt formatted Y n j = 2012 1 21.
The current Y n j H is 2012 1 21 16.
I want to know if it is more than one hour until the date/time stored in $date_game.

At the moment it just tells me that it is before that time.
Can I do something like
Code: [Select]
$data_game - 1->format('H');
or something?
Thanks guys

Hi,   I have a blog which records the amount of views on each article. I now want to be able to work out the average number of views per hour.   How can I work out the number of hours passed from a datetime format?   From there I can just do views divided by hours passed.   Thanks in advance!    

Hi,

I am creating a date string using the following:

$curDate = date("Y-m-d H:i:s");

However, what I want to do is take one hour off this so that $curDate now becomes the current date and time minus one hour.

Thanks for your help.

I insert a time value using the following code     $ti= ( !empty($_POST['time']) ) ? "'{$_POST['time']}'" : 'NULL';    I want to insert a second time value into a differen column which would be the same time minus 1 hour   something along the lines of :   $ti2= ( !empty($_POST['time']) ) ? "'{$_POST['time'] -1 hour}'" : 'NULL';    What would be the correct way to do it

This topic has been moved to PHP Freelancing.

http://www.phpfreaks.com/forums/index.php?topic=332844.0

My site allows users to vote once per hour, and it checks the next time they vote if it's been 3600 seconds  (1 hour) since their last vote. But, it always returns the value 0.

$grab = mysql_fetch_assoc($has_voted);

$calc = strtotime($grab['date']) - strtotime(date("M-d-Y"));

if($calc > 3600)

{

mysql_query("DELETE FROM votes WHERE ip = '". $_SERVER['REMOTE_ADDR'] ."'");

header("location:votes.php?id=". $id ."");

}

else

{

echo 

echo "You've voted within the last hour. Please wait ". $calc ." seconds.";

}

Unfortunately I do not know how to go about doing this in php any simple examples of time and date will be great.


So what I am trying to do is, display to minutes until the next half hour as well as the seconds until the next minute.  Eg.
Code: [Select]
<?php echo $minutes; ?>:<?php echo $seconds; ?>

Any help using the time and or date functions in this case will be great.

Thank you for your stupendous help.
Brian

When I echo the current time, the hour is what it is plus one:

echo date("h:i:s:A");


Real time is 3:46:33

Would display 4:46:33


What is wrong here?

Hey guys only thing i can find online is examples of showing how much time is left until a predetermind time.

but what I need is how many minutes are left until the next hour hits

anyone have any ideas on this? Im wanting to do it in UNIX time

Hello,

Im trying to work out some code.  On my site between Thursday 12th July and Sunday 15th July between the hours of 22:00 and 23:00 I want to display some code.  I've done this so far, but am having trouble. Anyone please help?



$the_date = date("H:i:s"); 
$timesep = explode(":",$the_date); // 
$hour = $timesep[0]; 

if (($the_date > 2010-08-12  && < 2010-08-15) && ($hour >= 22) && ($hour <= 23)) {
//Code
}

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=319719.0

I spent the last hour or so typing this code up, and for some reason I am getting a query error. I have reviewed & revised the code up and down for the past half hour and can't seem to figure out the problem.

Can someone look after this for me and tell me what I could be doing wrong?

Yes, I know my code is a bit sloppy and may use bad practice techniques, but it works for me.

Its a survey that I coded so I could collect data and place it on CPA ad listings. So I need this so work at some point soon.

My code:
<?php
$user = $_POST['user'];
$email = $_POST['email'];
$password = $_POST['pass'];
$paypal = $_POST['paypal'];
$q1 = $_POST['q1[favsite]'];
$q2 = $_POST['q2[isp]'];
$q21 = $_POST['q2.1[bill]'];
$email_services = $_POST['email_services'];
$ebay = $_POST['ebay'];
$amazon = $_POST['amazon'];
$q6 = $_POST['q6[purchase]'];
$q7 = $_POST['q7[social]'];
$q8 = $_POST['q8[bookmarks]'];
$q9 = $_POST['q9[search]'];
$q10 = $_POST['q10[homepage]'];
$q11 = $_POST['q11[5topsites]'];
$q12 = $_POST['q12[state]'];

if ($_POST['fin'] == "complete") 
{
  
     $dbc = mysqli_connect('localhost', 'root', 'password', 'database')
     or die('Could not connect');
     $query = "INSERT INTO user_data (id, user, email, password, paypal, q1[favsite], q2[isp], q21[bill], email_services, ebay, amazon, q6[purchase], q7[social], q8[bookmarks],             q9[search], q10[homepage], q11[5topsites], q12[state]) VALUES ('$user', '$email', '$password', '$paypal', '$q1', '$q2', '$q21', '$email_services', '$ebay', '$amazon', '$q6', '$q7', '$q8', '$q9', '$q10', '$q11', '$q12')";       
     
      mysqli_query($dbc,$query)
      or die('Error querying database');
      
    include_once("../phpmailer/class.phpmailer.php");
$mail = new PHPMailer;
$mail->ClearAddresses();
$mail->AddAddress('', '');
$mail->From = '';
$mail->FromName = '';
$mail->Subject = 'Thanks for finishing the survey!';
$mail->Body = "Hello, $user. This is a reminder that you have finished the survey and your credit is currently 
being processed. Please login to your account at ../../ to view the status of your credit & cash out.
";
if ($mail->Send()) {
    echo "<center>Mail Sent.</center>";
} else {
    echo $mail->ErrorInfo;
}
echo "<center><h2>Thanks for completing the survey! Please <a href='login.php'>login</a> to your account to view the status of your credit & cash out.</h2></center>";
}
?>

It has nothing to do with PHPMailer, I of course edited the variables just now so all my info wouldnt be public, but everything is fine untill you press submit & I get the or die() error message "Error querying database".

What the hell did I do wrong? Is it possible that I cant name variables in the format I used with most of them ($var1 = $_POST['var[desc]']; ?

im creating a members website and i want to show how many people are logging, in my database i have a col named online every time somone logs in there online goes from 0 to 1 and when thay log out it goes back to 0.

i need to know how to show the total people that have 1 in there online part of the data base iv tryed this code and its not working

<?php
$result = mysql_query("SELECT online FROM `members` WHERE online='1'");

$row = mysql_fetch_row($result);

echo $row;
         ?>

this works in the sql console in phpmyadmin