PHP - Need Help Configuring Store Hours

I am trying to make a function that is customizable to delete a query from a database.

Here's the function.
function checkHashs($hours) {
        if($getHashs = mysql_query('SELECT * FROM `hash_codes`')) {
          $dateQ = mysql_query('SELECT MAX(addedon) AS lastVoteDate FROM `hash_codes`') or die(mysql_error());
          $getDate = mysql_fetch_assoc($dateQ);
          $diffrence = time() - strtotime($getDate['lastVoteDate']);
  echo $diffrence / 60 / 60;
          if (($diffrence / 60 / 60) >= $hours) {
mysql_query('DELETE FROM `hash_codes` WHERE `addedon` = '. ($diffrence / 60 / 60) >= $hours .'') or die(mysql_error());
          }
}
}

And I am trying to call it by
checkHashs(24);

What I am trying to do is make it so when that function is ran it will delete all oh the guerys that have the `addedon` row over X hours(24 hours in this case).

Any ideas?

Hello

I am trying to do a cronjob to check if a user has been online within the past 24 hours.
I have a column with "last_login" with this format: "1337593284".

How can I do a simple if (last_login+24hours< servertime) {

do something...

Im using

$time = date('Y-m-d H:i:s');

How can I make a new variable that adds 4 hours on the $time output.

in my MySql database i have a field "created_at"  = now()
for ex: 2011-05-23 11:47:28

i need to add two hours to that so i have 2011-05-23 13:47:28
how will i do this please?

i have this so far but it is not correct:

    $currentTime = time($msg->getCreatedAt());
    $timeAfterOneHour = $currentTime+60*60;
    $date=date("Y-m-d H:i:s",$timeAfterOneHour);

please help
thank you

Hi, I have a database where I store times with the Now(); function.

What I want is:
When the viewer looks at the post, instead of it saying: Posted: 2010-11-14 19:08:14,
I want it to say posted: x hours ago, Posted: x days ago, etc.

Any help with this would be appreciated!

I would like to create function, what prints times to specified day.

For example

Day  time              total
1      10 - 12         2
2      0 (or nothing)
3      10 - 12:15    2:15
4      10:30 - 12    1:30
5      10 - 12         2
6      0 (or nothing)
7      0 (or nothing)

Time comes from database and there is column for day, starting time (like 10), ending time (like 12:15) and total, calculated, time in db. So day and times come from db.

Have thinked this much, but can't figure out how.

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

I have a weird issue.  I'm working on something where people will submit hours for a business.  Originally it was free form text, but then I figured it may get messy with people doing M-F 9-5 then Monday through Friday, 9AM to 5PM, Mon-Fri, etc.  So I switched to a bunch of inputs.  It has each day then a select for from and one for to.  Data is passed to the script as a form submission.  Vars are named HF# and HT# (hours from, hours to and a 0-6 digit for Mon-Sun).  I want to take the input and output a string as such:   Mon-Tue, 8:00 AM to 5:00 PM Thu-Fri, 8:00 AM to 5:00 PM Sat, 9:00 AM to 1:00 PM Sun, 12:00 PM to 3:00 PM   I did this:

$formdata = array();
$formdata["hf0"] = "8:00 AM";  $formdata["ht0"] = "4:30 PM";
$formdata["hf1"] = "8:00 AM";  $formdata["ht1"] = "4:30 PM";
$formdata["hf2"] = "";         $formdata["ht2"] = "";
$formdata["hf3"] = "8:00 AM";  $formdata["ht3"] = "4:30 PM";
$formdata["hf4"] = "8:00 AM";  $formdata["ht4"] = "4:30 PM";
$formdata["hf5"] = "9:00 AM";  $formdata["ht5"] = "1:00 PM";
$formdata["hf6"] = "12:00 PM"; $formdata["ht6"] = "3:00 PM";
 
$WeekDays = array("Mon", "Tue", "Wed", "Thu", "Fri", "Sat", "Sun");
$hours = ""; $sd = "Mon"; $ld = ""; $fr = $formdata["hf0"]; $to = $formdata["ht0"];
for ($x=0; $x<=6; $x++) {
  if (empty($sd) && !empty($formdata["hf$x"])) { $sd = $WeekDays[$x]; $fr = $formdata["hf$x"]; $to = $formdata["ht$x"]; }
  if ($fr != $formdata["hf$x"] || $to != $formdata["ht$x"]) {
    $hours .= "$sd" . (($sd==$ld)?", ":" - $ld, ") . " $fr to $to<br/>";
    if (!empty($formdata["hf$x"])) { $sd = $WeekDays[$x]; $fr = $formdata["hf$x"]; $to = $formdata["ht$x"]; } else { $sd = ""; }
  }
  $ld = $WeekDays[$x];
}

($formdata was because I was tired of hitting back, changing the data then resubmitting)  This kinda works, but it skips Sunday.  There has got to be a better way to do this..  any ideas?

$date = date('G:i', $row_teams['startTime']); // this date looks like this 15:30


I am trying to add 2 hours to it, any suggestions please

It was suppose to be my day off but here I set in front of the computer trying to figure out some code.

I can not get my code to send me the information that is inputted into the fields. The style is what I want but most importantly it has got to work. I'm going to put the PHP code and the HTML code. If you need the CSS code I can post that. I'm not sure if you guys need that though. Hopefully someone out there can help me out. I know it might be a very simple error causing me tons of problems. I just can't figure it out. I'm novice at this stuff.

PHP CODE & HTML CODE:

<?php
if ($_POST["email"]<>'') {
   $ToEmail = 'Justin@fullloop.net';
   $emailSubject = 'contact';
   $mailheader = "From: ".$_POST["email"]."\r\n";
   $mailheader .= "Reply-To: ".$_POST["email"]."\r\n";
   $mailheader .= "Content-type: text/html; charset=iso-8859-1\r\n";
   $MESSAGE_BODY = "Name: ".$_POST["name"]."<br>";
   $MESSAGE_BODY = "Phone: ".$_POST["phone"]."<br>";
   $MESSAGE_BODY .= "Your Email: ".$_POST["email"]."<br>";
   $MESSAGE_BODY .= "Your Message: ".nl2br($_POST["comment"])."<br>";
   mail($ToEmail, $EmailSubject, $MESSAGE_BODY, $mailheader) or die ("Failure");
?>
<html>
<body>

<?php echo $_POST["name"]; ?>,<br />
Thank you for contacting Full Loop Inspections. We take great pride in our services and we are grateful you are considering us to complete your inspection. An inspector will contact you at <?php echo $_POST["phone"]; ?> or email you at <?php echo $_POST["email"]; ?> .

</body>
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<?php
} else {
?>

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<?php
};
?>











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hii

I have made some code that works out the difference between two timecodes. IE

  Time 1                  Time 2         Time2-Time1
1330613232  -   1330613545 =        313

Im wanting to express the difference in terms of hours.

313 as a time code is 00:06:49 but I want it to show 0.12 hours.

how would I achieve this?

x

I have the below code wish I wish to add 12 hours to:

Code: [Select]
$today = date('D', mktime(0,0,0,date('m'),date('d')+0,date('Y')));
anyone to show how I add 12 hours?

Okay I want to run some code every 24 hours, what would be the best way to do this? Also I don't have root to the server or anything.

so i have a mysql column with a datetime and i was wondering how i could make it echo the time in "X minutes ago" or "X hours ago" and also, when its more then 24 hours old, delete the entire entry