turning YYYYMMDD to graphic calendar view

vinny199

Hi,

I'm very stuck with the following.
I am making "an event" website.
Events are inserted using a simple input calendar that posts as follow into MySql "data" table
$date = YYYYMMDD which in turns allows me to display my events ranked in the right order.

However, my display page is not that user friendly and I would like to generate a calendar view of all these dates.

I am not sure if this is possible or how to go about it.

Can you pls help?

Many thanks,

Vinny

vaaaska

You are not sure about which part - how to code it or if it's possible? It's certainly possible...

If you do a Google search on 'php calendar class' I bet you'll find a place to get your started. Otherwise, we have no idea where to really help you out.

Good luck...

vinny199

Ok, sorry
that was very vague, but yes, the first thing for me was to know whether this was possible.

Now I'll explain in more detail what I have, what i'd like to have and where I'm stuck

My table column containing the date is calle $b (which I know is not ideal, but I inherited that)

the table is called "job"

A typical date in the job table is $b = 2005/11/26

so far so good.
To display that in a more UK user friendly way on my "job page diary" I use:

<?php
setlocale(LC_TIME,fr_BE);
print strftime("%e %B %Y",strtotime($row_rsjob['b']));
?>

Which then displays "26 november 2005"
This is good in a way, but my friend complains he has to refer to his diary constantly to see what day it is etc.
I'm asked to improve this to a more graphical diary

The first thing would be to dispaly WHICH DAY the job is, like

Saturday 26 November 2005

So, my first question is, could you help me displaying the date is a way that shows the day of the week associated with the date?

My second question is:
in a ideal world, I'll get ot display the jobs underneath a Month header such as:

JANUARY
Monday 02 Janyary 2006
Tuesday 03 January 2006
Friday 07 January 2006

FEBRUARY
Thusday 11 February 2006

etc.

I hope this makes sense and you can help,

Many thanks,

Vinny

Installer

One way to do it:

Retrieve the date values as timestamps ("SELECT UNIX_TIMESTAMP()") from the database, and store them into an array (e.g. $dates). Then:

sort($dates);
setlocale(LC_TIME, fr_BE);
$indent = '&nbsp;&nbsp;';
$table_year = '';
$table_month = '';
foreach ($dates as $date) {
    $date_year = date('Y', $date);
    if ($date_year != $table_year) {
        $table_year = $date_year;
        $table_month = '';
        echo '<br />' . $table_year . '<br />';
    }
    $date_month = date('F', $date);
    if ($date_month != $table_month) {
        $table_month = $date_month;
        echo $indent . $table_month . '<br />';
    }
    echo $indent . $indent . strftime('%A %e %B %Y', $date) . '<br />';
    // Use "%#d" instead of "%e" for Windows machine
}

vinny199

Ok , thanks, but please be patient with me..
here is my sql statement:

SELECT *
FROM job
WHERE (UNIX_TIMESTAMP(TO_DAYS(B))  - TO_DAYS(NOW()) < 7 ) AND (TO_DAYS(B) - TO_DAYS(NOW()) >-1 ) AND  job.status='Confirmed'
ORDER BY job.b ASC ,  job.hours ASC ,  job.minutes ASC

(😎 is my date in the database formated as 2005/11/27

Where do I go from here before being able to use the code you kindly provided?

this staement produces jobs within 7 days.

Installer

I think this, or something close to it, will do what you want:

$query = "SELECT UNIX_TIMESTAMP(B) AS date FROM job
          WHERE (TO_DAYS(B) - TO_DAYS(NOW()) < 7) 
          AND (TO_DAYS(B) - TO_DAYS(NOW()) > -1) 
          AND job.status='Confirmed' 
          ORDER BY job.b ASC, job.hours ASC, job.minutes ASC";

$result = mysql_query($query);

setlocale(LC_TIME, fr_BE);
$indent = '&nbsp;&nbsp;';
$table_year = '';
$table_month = '';

while ($line = mysql_fetch_array($result)) {
    $date = $line['date'];
    $date_year = date('Y', $date);
    if ($date_year != $table_year) {
        $table_year = $date_year;
        $table_month = '';
        echo '<br />' . $table_year . '<br />';
    }
    $date_month = date('F', $date);
    if ($date_month != $table_month) {
        $table_month = $date_month;
        echo $indent . $table_month . '<br />';
    }
    echo $indent . $indent . strftime('%A %e %B %Y', $date) . '<br />';
    // Use "%#d" instead of "%e" for Windows machine
}

The "sort()" line is omitted because you're doing that in the query, and instead of putting the dates into an array, they can be handled inside the "mysql_fetch_array()" loop. And since you're only doing a week at a time you could probably do without the year heading b.s. I threw in. hth

vinny199

Thanks!
this works a treat.
I think I understand the idea of it too now

If i can bother you a little more...

Below each date listed, I have to list the basic details of the job and a little link to a "detailed page".

This is my code below.

There should be a way to integrate it underneath each date, shouldn't it?

   <table width="441" border="0" cellspacing="3" cellpadding="2">
                    <tr> 
                      <td width="16" valign="top">&nbsp;</td>
                      <td width="295"><p><strong><?php echo $row_rsjob['company']; ?></strong><br /> 
                          <?php echo $row_rsjob['name']; ?> <?php echo $row_rsjob['surname']; ?><br />
                        Job number: <strong><?php echo $row_rsjob['jid']; ?></strong><br /> 

                        </p>
                        <p><?php echo $row_rsjob['b']; ?><br /> 
                        </p></td>
                      <td width="106"><a href="jobmanagement2.php?<?php echo $MM_keepNone.(($MM_keepNone!="")?"&":"")."jid=".$row_rsjob['jid'] ?>">&gt;&gt; 
                        View details</a></td>
                    </tr>
                  </table>
                  <?php } while ($row_rsjob = mysql_fetch_assoc($rsjob)); ?>

Kudose

Just as a side note, I found this http://keithdevens.com/software/php_calendar and am currently using it here: http://www.mobilitycartrentals.com/index.php/calendar (see bottom for linked days)

Just thought I would pass it on because I found it very useful.