PHP - Translating Date To Week Number

I need to know which day of week was on particular date, so for example I would provide date lets say "Nov 22, 2010" formatted as needed and the function would return "Monday", is there a function in php to do that or how can it be done?

I have the week date range displaying the week range from sunday - saturday   $current_dayname = date("0"); // return sunday monday tuesday etc.
echo $date = date("Y-m-d",strtotime('last sunday')).'to'.date("Y-m-d",strtotime("next saturday"));   Which outputs in the following format 2015-01-25to2015-01-31   However, when I press next or previous, the dates don't change.   <?php
$year = (isset($_GET['year'])) ? $_GET['year'] : date("Y");
$week = (isset($_GET['week'])) ? $_GET['week'] : date('W');
if($week > 52) {
    $year++;
    $week = 1;
} elseif($week < 1) {
    $year--;
    $week = 52;
}
?>

<a href="<?php echo $_SERVER['PHP_SELF'].'?week='.($week == 52 ? 1 : 1 + $week).'&year='.($week == 52 ? 1 + $year : $year); ?>">Next  Week</a> <!--Next week-->
<a href="<?php echo $_SERVER['PHP_SELF'].'?week='.($week == 1 ? 52 : $week -1).'&year='.($week == 1 ? $year - 1 : $year); ?>">Pre Week</a> <!--Previous week-->

<table border="1px">
    <tr>
        <td>user</td>
<?php
if($week < 10) {
    $week = '0'. $week;
}
for($day= 1; $day <= 7; $day++) {
    $d = strtotime($year ."W". $week . $day);

    echo "<td>". date('l', $d) ."<br>". date('d M', $d) ."</td>";
}
?>
    </tr>
</table>

Hi,

I have a job listing website which displays the closing date of applications using:

$expired_date

(This displays a date such as 31st December 2019)

I am trying to show a countdown/number of days left until the closing date.

I have put this together, but I can't get it to show the number of days.

                    <?php 
                    $expired_date = get_post_meta( $post->ID, '_job_expires', true );
                    $hide_expiration = get_post_meta( $post->ID, '_hide_expiration', true );
                    if(empty($hide_expiration )) {
                        if(!empty($expired_date)) { ?>
                                <span><?php echo date_i18n( get_option( 'date_format' ), strtotime( get_post_meta( $post->ID, '_job_expires', true ) ) ) ?></span>
                    
                    <?php
                                        $datetime1 = new DateTime($expired_date);
                                        $datetime2 = date('d');
                                        $interval = $datetime1->diff($datetime2);
                                        echo $interval->d;

                     ?>
                        <?php }
                    } 
                    ?>

Can anyone help me with what I have wrong?

Many thanks

Hi, I am trying to get the number of days between the current date and a date in the future specified by column 'end_date'. The code I have seems to be working but it displays the number of days as a negative number, how do I change this to be a positive number? I have tried simply changing $days = $now - $end_date; to $days = $end_date - $now; but that doesn't work as I thought it would! Thanks in advance..

Code: [Select]
$now = time();
      $end_date = strtotime($row['end_date']);
      $days = $now - $end_date;
      echo floor($days/(60*60*24));

I want to set a never expire date, so I figured I would set the date 99 years into the future.  No dice.  After trial and error, it appears you can only make it 25 years into futu

date("Y-m-d",strtotime("+25 years"));

I wonder if it is the absolute date, or the number 25.  If it is the maximum date (i.e. nothing after 2035) then I will have a problem with this code next year!

Hi guys

I'm kinda stuck..so hopefully you guys can lend me a hand

I've got an array containing date elements ("Y-m-d")...

I'm trying to output some data into googlecharts..so i need to count how many elements that are in the array with todays date -1 day, todays date-2 days...todays date -3days etc... Up until a set number of days (for example 7 days, 14 days etc)..

Any advice on how to go about to achieve this?

Hoping someone can help me on this, but I had some help on getting this code (or part of it) to finally work the way I wanted it, however, there are a few end-user issues with it now.

What this code does is disables the user(s) from posting any kind of links (www.example.com or example.com). However, now I'm trying to....expand it a little I guess.

Here is the current code:
Code: [Select]
if (stristr($pagetext, 'http://') OR stristr($pagetext, 'www.') OR stristr($pagetext,'@') OR stristr($pagetext, '[URL') OR stristr($pagetext, '[url') OR stristr($pagetext, '[IMG') OR stristr($pagetext, '[img') OR preg_match("#[a-z0-9]([-a-z0-9]+)?(\.[a-z]{2,3})?(\.[a-z]{2,4})#i", $pagetext))
            {
                more code here
            }
Now, I KNOW what most of that code does, but what I need, is what exactly this code does:
Code: [Select]
preg_match("#[a-z0-9]([-a-z0-9]+)?(\.[a-z]{2,3})?(\.[a-z]{2,4})#i", $pagetext)Can anyone break this down and tell me what exactly does what?

Thanks in advance

I'm getting the dreaded " Invalid parameter number: number of bound variables does not match number of tokens" error and I've looked at this for days.  Here is what my table looks like:

|  id          | int(4)       | NO   | PRI | NULL    | auto_increment |
| user_id     | int(4)       | NO   |     | NULL    |                |
| recipient   | varchar(30)  | NO   |     | NULL    |                |
| subject     | varchar(25)  | YES  |     | NULL    |                |
| cc_email    | varchar(30)  | YES  |     | NULL    |                |
| reply       | varchar(20)  | YES  |     | NULL    |                |
| location    | varchar(50)  | YES  |     | NULL    |                |
| stationery  | varchar(40)  | YES  |     | NULL    |                |
| ink_color   | varchar(12)  | YES  |     | NULL    |                |
| fontchosen  | varchar(30)  | YES  |     | NULL    |                |
| message     | varchar(500) | NO   |     | NULL    |                |
| attachment  | varchar(40)  | YES  |     | NULL    |                |
| messageDate | datetime     | YES  |     | NULL    |        

Here are my params:

            $params = array(                 
               ':user_id' => $userid,                 
               ':recipient' => $this->message_vars['recipient'],                 
               ':subject' => $this->message_vars['subject'],                 
               ':cc_email' => $this->message_vars['cc_email'],                 
               ':reply' => $this->message_vars['reply'],                 
               ':location' => $this->message_vars['location'],                 
               ':stationery' => $this->message_vars['stationery'],                 
               ':ink_color' => $this->message_vars['ink_color'],                 
               ':fontchosen' => $this->message_vars['fontchosen'],                 
               ':message' => $messageInput,                 
               ':attachment' => $this->message_vars['attachment'],                 
               ':messageDate' => $date             );

Here is my sql:

$sql = "INSERT INTO messages (user_id,recipient, subject, cc_email, reply, location,stationery, ink_color, fontchosen, message,attachment) VALUES( $userid, :recipient, :subject, :cc_email, :reply, :location, :stationery, :ink_color, :fontchosen, $messageInput, :attachment, $date);";

And lastly, here is how I am calling it:
            $dbh = parent::$dbh;             
           $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_WARNING);              
           if (empty($dbh))    return false;             
                  $stmt = $dbh->prepare($sql);             
                 $stmt->execute($params) or die(print_r($stmt->errorInfo(), true));              
           if (!$stmt) {    print_r($dbh->errorInfo());              }

I know my userid is valid and and the date is set above (I've echo'd these out to make sure).  Since the id is auto_increment, I do not put that in my sql (though I've tried that too), nor in my params (tried that too).  What am I missing?  I feel certain it is something small, but I have spent days checking commas, semi-colons and spelling.  Can anyone see what I'm doing wrong?

I'm helping a friend in developing a project for our alma mater. And its based on this paper: An Enhanced Indoor Positioning Method Based on Wi-Fi RSS Fingerprinting

So far 80% of the web app is finished except for the algorithm (We have no experience with algorithms to source code translations) that needs to be implemented in the website. Inside this paper there is this algorithm that the proponents of the research used. (attached as screenshot).

The algorithm was converted to code by using MATLAB, but we are trying to use PHP with our project. Based on the pseudo code the calculation of the Euclidean Distance is the core of it. Can anyone help us with the translation just to get it started. 

 

 

Hello there,

I have a DB table where everyday I insert my meetings for the future.

I want to ask you guys if anybody can help me with creating some sort of a weekly report.

EXAMPLE:

I have meeting from the beginning of the year (Mon-Sat)...could be 4-10 meeting per day, so a lot in general.

I want them to appear , with sql, under the WEEK they were in for example:
(a year has 52 weeks)

WEEK 1 (01-01-2011 -> 08-01-2011)
*M1
*M2
*M3
...

WEEK 2 (09-01-2011 -> 16-01-2011)
*M4
*M5
*M6
...


etc...

Is there a way to define the date's range after the week ended automatically?

Thanks

Hello All,

I have read through this script a million times, and I can not for the life of me figure out how the script decides which day of the week to start on. Right now the week starts on Monday (which goes against every other calendar we use). I need to figure out how to make the week start on Sunday.

Here's the script:
<style type='text/css'>
.outofrange
{
background-color: #d9cdb3;
}
.currentday
{
background-color: #ccc;
}
</style>

<?PHP
$Year = preg_replace("/[^0-9]/","",$_GET['year']);
$Month = preg_replace("/[^0-9]/","",$_GET['month']);

if ($Month == "") { $Month = date("n"); }
if ($Year == "") { $Year = date("Y"); }
if ($Month==0)  { $Month=12; $Year--; }
if ($Month==13) { $Month=1;  $Year++; }

$myServer = "xxxxxx";
$myUser = "xxxxxx";
$myPass = "xxxxxx";
$myDB = "Calendar"; 

//connection to the database
$dbhandle = mssql_connect($myServer, $myUser, $myPass)
  or die("Couldn't connect to SQL Server on $myServer"); 

//select a database to work with
$selected = mssql_select_db($myDB, $dbhandle)
  or die("Couldn't open database $myDB"); 

//declare the SQL statement that will query the database
$query = "
SELECT *
FROM calendar
WHERE school = '00'
";

//execute the SQL query and return records
$result = mssql_query($query);

//display the results 


while($row = mssql_fetch_array($result))
{
if (date("Y",strtotime($row['date']))==$Year)
{
$Event[date("z",strtotime($row['date'])+1)+1] .= $row['event'] . "<br />" . "<font color='#" . $row['colors'] . "'>" .  $row['location']  . "</font>" . "<p>";
}
}


$Timestamp = strtotime("$Year-$Month-01");

for ($Day=1;$Day<=32;$Day++)
{
if (checkdate($Month,$Day,$Year))
{
$LastDayInMonth = $Day;
}
}

$FirstDayInMonth = date("N",$Timestamp);

for ($MakeCal=$FirstDayInMonth;$MakeCal<=$FirstDayInMonth+$LastDayInMonth;$MakeCal++)
{
//echo "$MakeCal<br />";
}
$WeeksNeeded = ceil(($FirstDayInMonth+$LastDayInMonth)/7);

echo "<div align='center'><span class='bodyb'>" . date("F Y",$Timestamp) . "</span></div><p>";

echo "<table width='100%' border='0'> 
<tr>
<td align='left' valign='top'><a href='index.php?month=" . (($Month)-1) . "&amp;year=$Year' class='monthview'>Previous Month</a></td>
<td align='right' valign='top'><a href='index.php?month=" . (($Month)+1) . "&amp;year=$Year' class='monthview'>Next Month</a></span></td>
</tr>
</table>";
echo "<p>";

//Main Table starting with days of week through the month views//

echo "<table width='615px' cellspacing='0' cellpadding='2' border='1' bordercolor='#666666'>
<tr>
<th>Monday</th>
<th>Tuesday</th>
<th>Wednesday</th>
<th>Thursday</th>
<th>Friday</th>
<th>Saturday</th>
<th>Sunday</th>
</tr>
";

for ($DoWeeks=1;$DoWeeks<=$WeeksNeeded;$DoWeeks++)
{
for ($DoDays=1;$DoDays<=7;$DoDays++)
{
$CurrentDay++;
if ($CurrentDay < $FirstDayInMonth)
{
echo "<td width='10%' class='outofrange'>&nbsp;</td>";
}
else
{
if (($CurrentDay-$FirstDayInMonth)+1>$LastDayInMonth)
{
echo "<td width='10%' class='outofrange'>&nbsp;</td>";
}
else
{
$DayOfYear = date("z",strtotime("$Year-$Month-" . (($CurrentDay-$FirstDayInMonth)+1))) + 1;
if ($DayOfYear==date("z")+1 && date("Y")==$Year)
{
echo "<td width='10%' height='100' valign='top' class='currentday'><strong>" . (($CurrentDay-$FirstDayInMonth)+1) . "</strong><br />";
}
else
{
echo "<td width='10%' height='100' valign='top'><strong>" . (($CurrentDay-$FirstDayInMonth)+1) . "</strong><br />";
}

if ($Event[$DayOfYear]!="")
{
echo $Event[$DayOfYear] . $Location[$DayOfYear] . "<p /></font>";
}
echo "</td>";
}
}
}
echo "</tr><tr>";
}
echo "</table><p>";

I'm trying to produce a weekly calendar. I can't figure how to go about getting the sundays of particular weeks of the year. I know you can get the week number( date('W') produces 1 - 52) of a particular date but i can't really figure how to use that number to get the sunday in that week. Anybody have any suggestions?

It would be nice if i could write a function where i input the week number and the year and it returns the sunday of that week. Can't figure out how to do that though.

Hi Guys,

I'm new to PHP. I've gone through a few tutorials and I'm not sure how to go about the following:

For work, every week 1 of 3 people is responsible for checking some bug report. This week I have to do this.
Next week it's Tom the week after that is Brenda. Then it's back to me again and it cycles. We currently have a white board calendar at the back of the office that gets updated with the name of the person responsible for doing the checks that week.

What I'm trying to do is write a small PHP snippet that will show on our internal site something like:

Bug Checker this week is: TARA
Then the next week it'll show Tom, then Brenda. And cycle back. That way we don't have to keep updating that board. Does anybody know the best way to go about this. I keep thinking of different ideas but they all seem pretty convoluted and I'm fairly certain there's got to be a very easy way to do it; it's just not coming to me.


Any help would be appreciated  Thanks so much!

XXX
Tara

EDIT: We cycle through every Monday. If that helps.

I considered creating a database file with the date and the person and just read that in but that is so embarassingly inefficient that I didn't even want to mention this in the post.

 

I wish to find the closest two DateTimes which are within $fillStart and $fillEnd and have the same week and 24 hour times as $gapStart and $gapEnd.

For instance, the following results in $fillStartModified and $fillEndModified which meets that criteria.

fillStart           2019-07-23 00:15:00 Tue
fillEnd             2019-09-23 13:00:00 Mon
gapStart            2019-05-23 00:15:00 Thu
gapEnd              2019-06-23 13:00:00 Sun
fillStartModified   2019-07-25 00:15:00 Thu
fillEndModified     2019-08-25 13:00:00 Sun

I came up with the following which seems to work, but it is kind of complicated and I am not positive it will meet all edge conditions.  Any recommendations?  Thanks

function getFill(\DateTimeImmutable $fillStart, \DateTimeImmutable $fillEnd, \DateTimeInterface $gapStart, \DateTimeInterface $gapEnd)
{
    //$gapInterval = $gapStart->diff($gapEnd);  // Doesn't work 
    $gapInterval = new \DateInterval('PT'.($gapEnd->getTimestamp() - $gapStart->getTimestamp()).'S');
    if($gapStart > $fillStart) {
        //The fill is older than the gap so make the fill's endTime match the gap's endTime
        $fillEndModified = $fillEnd
        ->modify('previous '.$gapEnd->format('D'))
        ->setTime(...explode('.', $gapEnd->format('H.i.s.u')));
        if($fillEndModified->diff($fillEnd)->format('%a') >= 7) {
            $fillEndModified = $fillEndModified->add(new \DateInterval('P7D'));
        }
        $fillStartModified = $fillEndModified->sub($gapInterval);
        if($fillStartModified < $fillStart) {
            $fillStartModified=null;
            $fillEndModified=null;
        }
    }
    else {
        //The fill is newer than the gap so make the fill's startTime match the gap's startTime
        $fillStartModified = $fillStart
        ->modify('next '.$gapStart->format('D'))
        ->setTime(...explode('.', $gapStart->format('H.i.s.u')));
        if($fillStart->diff($fillStartModified)->format('%a') >= 7) {
            $fillStartModified = $fillStartModified->sub(new \DateInterval('P7D'));
        }
        $fillEndModified = $fillStartModified->add($gapInterval);
        if($fillEndModified > $fillEnd) {
            $fillStartModified=null;
            $fillEndModified=null;
        }
    }
    return [$fillStartModified, $fillEndModified];
}