Greg, That worked brilliantly... I bow down to your sheer excellence!!! Thankyou Thankyou Thankyou
"Greg Wiley" <[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] Funny you should ask this because I've been meaning to share some code that I wrote a few months ago that does exactly this for any day of the week. Please feel free to use this and make changes. I'd appreciate any changes/bug fixes being sent back to me though. <?php /* * num_days * * A function that takes a day, a month and a year and returns the number of * occurrences of that day in the given month and year. * * Arguments: * day - the day required * month - which month are we talking about * year - which year are we talking about * * Returns: * occ - the number of times the day occurs in the month. * or NULL if there are invalid parameters. */ function num_days ($day, $month, $year) { $day_array = array("Mon" => "Monday", "Tue" => "Tuesday", "Wed" => "Wednesday", "Thu" => "Thursday", "Fri" => "Friday", "Sat" => "Saturday", "Sun" => "Sunday"); $month_array = array(1 => "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec"); /* * Check our arguments are valid. */ /* * $day must be either a full day string or the 3 letter abbreviation. */ if (!(in_array($day, $day_array) || array_key_exists($day, $day_array))) { echo "num_days: invalid argument. \$day must be day name or three letter abbreviation"; return; } /* * $month must be either a full month name or its 3 letter abrreviation */ if (($mth = array_search(substr($month,0,3), $month_array)) <= 0) { echo "num_days: invalid argument. \$month must be month name or three letter abbreviation"; return; } /* * Now fetch the previous $day of $month+1 in $year; * this will give us the last $day of $month. */ /* * Calculate the timestamp of the 01/$mth+1/$year. */ $time = mktime(0,0,0,$mth+1,1,$year); $str = strtotime("last $day", $time); /* * Return nth day of month. */ $date = date("j", $str); /* * If the difference between $date1 and $date2 is 28 then * there are 5 occurences of $day in $month/$year, otherwise * there are just 4. */ if ($date <= 28) { return 4; } else { return 5; } } ?> The other function I have does this: /* * nth_day * * A function that takes a number and a day and returns the date of * nth occurrence of that day in the given month and year. * * Arguments: * nth - the nth occurence required * day - the day required * month - which month are we talking about * year - which year are we talking about * * Returns: * date - the date on which the nth day occurs in month and year. * or NULL for errors. */ Give me a shout if anyone wants this one. Cheers, Greg. > -----Original Message----- > From: Phil Dowson [mailto:[EMAIL PROTECTED] > Sent: 29 September, 2003 16:22 > To: [EMAIL PROTECTED] > Subject: [PHP] How many Mondays in a month? > > > Hi, > > I have posted a similar question in php.db, but I was > wondering if someone could help out with a non-db issue. I am > trying to display statistics of visitors to my web site, and > what I would like to do is show the average number of > visitors that have visited the site in a given month for a > certain day e.g..: > > Stats for www.mysite.com for 09/2003 > > Monday - 15 average - 65 total > Tuesday - 16 average - 66 total > Wednesday - 14 average - 65 total > Thursday - 13 average - 63 total > Friday - 15 average - 65 total > Saturday - 5 average - 25 total > Sunday - 6 average - 28 total > > I have tried a number of ways to do this, but I cannot work > out a way to show the number for example Mondays that will be > in a given month, which I need to work out the average. I can > work out the total easy enough, just not the average. > > TIA > > Phil Dowson > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > > -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
