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    <p><strong>PHP</strong></p> <pre><code>$week_number = 40; $year = 2008; for($day=1; $day&lt;=7; $day++) { echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n"; } </code></pre> <p><hr> Below post was because I was an idiot who didn't read the question properly, but will get the dates in a week starting from Monday, given the date, not the week number..</p> <p><strong>In PHP</strong>, adapted from <a href="http://ie2.php.net/manual/en/function.date.php#85258" rel="noreferrer">this post</a> on the <a href="http://ie2.php.net/manual/en/function.date.php" rel="noreferrer">PHP date manual page</a>:</p> <pre><code>function week_from_monday($date) { // Assuming $date is in format DD-MM-YYYY list($day, $month, $year) = explode("-", $_REQUEST["date"]); // Get the weekday of the given date $wkday = date('l',mktime('0','0','0', $month, $day, $year)); switch($wkday) { case 'Monday': $numDaysToMon = 0; break; case 'Tuesday': $numDaysToMon = 1; break; case 'Wednesday': $numDaysToMon = 2; break; case 'Thursday': $numDaysToMon = 3; break; case 'Friday': $numDaysToMon = 4; break; case 'Saturday': $numDaysToMon = 5; break; case 'Sunday': $numDaysToMon = 6; break; } // Timestamp of the monday for that week $monday = mktime('0','0','0', $month, $day-$numDaysToMon, $year); $seconds_in_a_day = 86400; // Get date for 7 days from Monday (inclusive) for($i=0; $i&lt;7; $i++) { $dates[$i] = date('Y-m-d',$monday+($seconds_in_a_day*$i)); } return $dates; } </code></pre> <p>Output from <code>week_from_monday('07-10-2008')</code> gives:</p> <pre><code>Array ( [0] =&gt; 2008-10-06 [1] =&gt; 2008-10-07 [2] =&gt; 2008-10-08 [3] =&gt; 2008-10-09 [4] =&gt; 2008-10-10 [5] =&gt; 2008-10-11 [6] =&gt; 2008-10-12 ) </code></pre>
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    1. POCalculating days of week given a week number
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    1. COThat input is not as specified. Input would be, for example, "41".
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    2. COThanks, bad case of Monday morning brain here, didn't read it properly. Answer updated with much shorter code snippet!
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    3. COthat code snippet is not working for me. You don't use $year ...
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