# In PHP, how to know how many mondays have passed in this month uptil today?

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Assume today is Feb 21, 2011 ( Monday ). It is the third Monday of this month. If date is given as input, How can I know how many Mondays have passed before it?

In PHP, how to know how many mondays have passed in this month uptil today?

Solution :

``````\$now=time() + 86400;
if ((\$dow = date('w', \$now)) == 0) \$dow = 7;
\$begin = \$now - (86400 * (\$dow-1));

echo "Mondays: ".ceil(date('d', \$begin) / 7)."<br/>";
``````

works for me….

EDIT: includes today’s monday too

That sounds like a pretty straightforward division calculation. From the current date, subtract number of days past last monday (example: wednesday = -2), divide it by 7 and `ceil()` it to round it up.

EDIT: That will include the current monday in the number, returning “3” for monday 21st.

``````    <?php

function mondays_get(\$month, \$stop_if_today = true) {

\$timestamp_now = time();

for(\$a = 1; \$a < 32; \$a++) {

\$day = strlen(\$a) == 1 ? "0".\$a : \$a;
\$timestamp = strtotime(\$month . "-\$day");
\$day_code = date("w", \$timestamp);
if(\$timestamp > \$timestamp_now)
break;
if(\$day_code == 1)
@\$mondays++;

}

return \$mondays;
}

echo mondays_get('2011-02');
``````

Hope this is of use to you! i’ve just rolled it up.

“Beware of bugs in the above code; I have only proved it correct, not tried it.”

Works OK afaik

You could loop through all the days until now and count the mondays:

``````\$firstDate = mktime(0, 0, 0, date("n"), 1, date("Y"));
\$now = time();
\$mondays = 0;
for (\$i = \$firstDate; \$i < \$now; \$i = \$i + 24*3600) {
if (date("D", \$i) == "Mon")
\$mondays ++;
}
``````

Haven’t tested this script

Try this…

``````//find the most recent monday (doesn't find today if today is Monday though)
\$startDate = strtotime( 'last monday' );

//if 'last monday' was not this month, 0 mondays.
\$mondays = date( 'm', \$startDate ) != date( 'm' )
? 0
: floor( date( 'd', \$startDate ) / 7 );

//increment the count if today is a monday (since strtotime didn't find it)
if ( date( 'w' ) == 1 ) \$mondays++;
``````

Another way is to find what day of the week is today, find the first such day of the month via some magic `strtotime()`, then calculate the difference between that and now in weeks. See below for a function that will take a `Y-m-d` formatted `date()` and return which weekday of the month it is.

Note: `strtotime` needs to be verbose, including “of” and the month: “first Monday of 2011-02” otherwise it advances one day. This bit me when I was testing edge cases.

Also added some display pepper which is completely optional but I felt like it.

``````function nthWeekdayOfMonth(\$day) {

\$dayTS = strtotime(\$day) ;

\$dayOfWeekToday = date('l', \$dayTS) ;

\$firstOfMonth = date('Y-m', \$dayTS) . "-01" ;
\$firstOfMonthTS = strtotime(\$firstOfMonth) ;

\$firstWhat = date('Y-m-d', strtotime("first \$dayOfWeekToday of \$monthYear", \$firstOfMonthTS)) ;
\$firstWhatTS = strtotime(\$firstWhat) ;

\$diffTS = \$dayTS - \$firstWhatTS ;
\$diffWeeks = \$diffTS / (86400 * 7);

\$nthWeekdayOfMonth = \$diffWeeks + 1;

return \$nthWeekdayOfMonth ;
}

\$day = date('Y-m-d') ;
\$nthWeekdayOfMonth = nthWeekdayOfMonth(\$day) ;

switch (\$nthWeekdayOfMonth) {
case 1:
\$inflector = "st" ;
break ;
case 2:
\$inflector = "nd" ;
break ;
case 3:
\$inflector = "rd" ;
break ;
default:
\$inflector = "th" ;
}

\$dayTS = strtotime(\$day) ;

\$monthName = date('F', \$dayTS) ;
\$dayOfWeekToday = date('l', \$dayTS) ;

echo "Today is the {\$nthWeekdayOfMonth}\$inflector \$dayOfWeekToday in \$monthName" ;
``````