[SOLVED] Looping puzzle involving dates.

[Maracles]

I have a for loop which creates a table consisting of 28 rows:

 

	for ($i=1; $i < 29; $i++) {
		echo '<td>'.$i.'</td>';
		echo '<td>'.$newdate.'</td>';
		echo '<td>'.$days[$b++].'</td>';
		echo '</tr>';
		}

 

The first column is simply a number i.e. '1, 2' so forth.

The second column is a date.

The third column is the day i.e. Mon, Tues Wed.

 

My problem is that I need the date in the date column to increase by one day every time the loop runs, however the first date in the first row must be the same as a variable I have in my database called $strtdate. Whilst I have read how to increase a date by one day I do not know how to do this in a loop where the first date is set by a variable.

 

Furthermore I need the day value in the adjacent column to correspond with the day (currently I am using a silly array just as a filler).

 

Any help would be much appreciated.

[DavidAM]

Assign $strtdate to $newdate BEFORE the loop.

 

Inside the loop:

Use the PHP date function to get the day of the week: $DOW = date('D', $newdate);

Increment $newdate by 1 day

[Maracles]

Thanks for the reply but i'm not quite sure what you mean. I have just tried to incorporate the $DOW part of the code and get the following error:

 

'Notice: A non well formed numeric value encountered in C:\wamp\www\Sandbox\cxapp\stckprofile.php on line 148'

 

Also, what way would recommend for incrementing the date, having read online there seems to a number of alternate ways.

 

 

[DavidAM]

I guess I should have asked: What is actually in $strtdate?  You say it is a column in the database, what is the data type of the column?  If it is a mySql DateTime, then it needs to be converted to a unix timestamp.  You can do that in the original query using "SELECT UNIX_TIMESTAMP(date_column_name_here) ..." or use strtotime() in PHP.

 

Then the $DOW code should work.  Note: 'D' in there returns the three character abbreviation, for the long name use 'l' (that's a lower-case L)

 

To increment a unix timestamp add one-day's worth of seconds ( 24 * 60 * 60 )

 

   $newdate = strtotime($strtdate);
   for ($i=1; $i < 29; $i++) {
         echo '<td>'.$i.'</td>';
         // Oh yeah, now it's an integer, so we want for format it
         echo '<td>'. date('m/d/Y', $newdate) .'</td>';
         //echo '<td>'.$days[$b++].'</td>';
         echo '<td>'. date('D', $newdate) . '</td>';
         echo '</tr>';
         $newdate += (24 * 60 * 60);
         }

[Maracles]

Thank you very much, I have just added that code and it works perfectly. I've been trying to figure that out for a couple of days so its a big help.

 

I have really learnt too much about dates and times yet and exactly how to calculate and formulate them but wanted to get this sorted before I got that far.

 

 

Will mark solved.

[DavidAM]

Cool glad to help!  And thanks for marking it solved.  That helps the next guy.

 

Dates are a pain.  Every environment handles them different.  In unix (and therefore) PHP they are integers, in mySql they look like strings but are always displayed in a strict format and you HAVE TO insert/update/query them in that format.  Some functions apply your TimeZone and some apply the server's timezone, and some just ignore you (they're called blind-dates -- Just Kidding).  When you start working with dates in a particular page pick one or the other (PHP's or mySql's) and be consistent.  Unix dates are easy to work with, because they are integers.  But watch out, day 1 is defined as Jan 1 1970 (or is that Day Zero?).  So, you can't event show my birthday in PHP (well, I think PHP 5 now supports negative dates). ... 

 

Ok, I'll stop ranting now.      I need a drink.