PHP - Broadcast Calendar Week Number

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

This topic has been moved to Linux.

http://www.phpfreaks.com/forums/index.php?topic=309899.0

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.

 

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];
}

 

I have a PHP script that downloads a file from a remote server but in order to download the correct file it has to subtract 1 day of the week.  It used to work quite some time ago so I dont know what is going on now.  The section of the script goes like this....

$days = array("sun","mon","tue","wed","thu","fri","sat");
$today = date("w");
$yesterday = $days[$today - 1];

Then it joins the 3 letter version of the day of week with a file name to grab the correct file download.  $yesterday is not returning the correct value instead I am getting an error...

# ./update.php 
PHP Notice:  Undefined offset: -1 in /root/fcc/work/update.php on line 6
fetching l_am_.zip
ncftpget: server said: l_am_.zip: No such file or directory.
PHP Warning:  unlink(counts): No such file or directory in /root/fcc/work/update.php on line 23
data set empty nothing to process 

Sun, 01 February, 2015 6:29:06 PM

So in the above error I can see that $yesterday is "Undefined offset" and because of that the file it tries to download is incorrect, it should have tried to download l_am_sat.zip and it left out the 3 letter date code.
Edited by chadrt, 01 February 2015 - 09:48 PM.

Hi,

I'm trying to alter the code below. It cuurently finds the start and end date of this week, no matter what day you're on.

What I now need is to find the start & end date of next week, again, no matter what day you're on of this week

Thanks to aleX_hill for the following code:

<?php
#THIS WEEK
$dayOfWeek = date("w"); #The number of the day of the week. Sun = 0, Sat = 1
$todayDayOfMonth = date("d"); #The day of the month
$thisMonth = date("m"); #The number of this month. Jan = 1, Dec = 12

$firstDayOfWeek = $todayDayOfMonth - $dayOfWeek; #The day of the month for the beginning of the week (Sun)
if($firstDayOfWeek < 1) #The beginning of the week was in the previous month
{
$monthBeginWeek = $thisMonth - 1;
$firstDayOfWeek = $firstDayOfWeek + date("t",mktime(0,0,0,$monthBeginWeek,1,date("Y")));
} else {
$monthBeginWeek = $thisMonth;
}
$lastDayOfWeek = $todayDayOfMonth + (6 - $dayOfWeek); #Get the day of the month of the end of the week (sat)
if($lastDayOfWeek > date("t")) #if the day of the month is larger then the number of days in the month
{
$monthEndWeek = $thisMonth + 1;
$lastDayOfWeek = $lastDayOfWeek - date("t"); #Then take the number of days in the week to get the day number of the next week
} else {
$monthEndWeek = $thisMonth;
}
$start_of_week = date("Y-m-d",mktime(0,0,0,$monthBeginWeek,$firstDayOfWeek,date("Y")));
#echo $start_of_week;
$end_of_week = date("Y-m-d",mktime(0,0,0,$monthEndWeek,$lastDayOfWeek,date("Y")));
#echo $end_of_week;
?>



Any help is much appreciated!

TIA

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.

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>";

Hello, I have problem durring binding update query. I can't find what is causing problem.

public function Update(Entry $e)
    {
        try
        {
            $query = "update entry set string = $e->string,delimiter=$e->delimiter where entryid= $e->id";
            $stmt = $this->db->mysqli->prepare($query);
            $stmt->bind_param('ssi',$e->string,$e->delimiter,$e->id);

            $stmt->close();
           

        }
        catch(Exception $ex)
        {
            print 'Error: ' .$ex->getMessage();
        }
    }

When I run function update I'm getting next error:Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement   Can you help me to solve this problem ?
Edited by danchi, 17 October 2014 - 10:25 AM.

I need to display a number(the number is retrieved from the db) in the form input field such that only the last 4 digits is visbile, the remaining can be masked as * or X or whatever is applicable.

I know the last 4 can be obtained as follows:

    Code: [Select]
$number=substr($number,-4,4);
 
But when i hit the submit button the form validates the input field and checks if the number is a valid number of a specific format.
Therefore when I click on the submit button then I should still be able to unmask the masked numbers or do something similar that would help me validate the whole number.

  Code: [Select]
  <input type="text" name="no" value="<?php if(!empty($number)){ echo $number;} ?>">

I want to have the  timestamps  of  week numbers, this is working perfectly from the year 2011 to  2012.  It refuses to switch to the year 2013. 
Do anyone know what i'm doing wrong?:

<?php

$year='2012';
$weekno='01';
$date = strtotime($year.'W'.$weekno);

for($i=0;$i<55;$i++){
        print date('d-m-Y W',$date)."\n";
        $date=(weekno($date,'1'))."\n";
        $date=(int)$date;
}


function weekno($date,$step=''){
        $year=date('Y',$date);
       $weekno=date('W',$date);
        $date = strtotime($year.'W'.$weekno);
        if($step>0){
                $date = strtotime($year.'W'.$weekno. '+1 week');
        }
        if($step<0){
                $date = strtotime($year.'W'.$weekno. '-1 week');
        }
       return($date);

}


?>

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=329323.0

I've got a table with 5 columns that shows appointments for Monday through Friday, but I'm using 5 separate queries, one for each column, for each corresponding day because i cannot figure out how to query the whole week and only put the appointments in the column they belong. Thank you for any help.