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Calculate Number of Days, Excluding Sundays, Between Two Dates

Posted on 2008-06-19
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I'm using this code to calculate the difference (number of days) between any two dates.
Both \$begin_mk and \$end_mk are in YYYY-mm-dd format, like   2008-06-19 .

How could I modify it to exclude Sundays from the calculation?

For example, if there are 90 calendar days from 2008-01-01 to 2008-03-31 , then I guess I need to:
1) calculate the number of Sundays between those two dates, and
2) subtract the number of Sundays from the total number of days (90, in this case)

How to calculate the number of Sundays between any two dates? Or is there some better, easier way to do this?
``````\$begin_mk = strtotime("\$begin");
\$end_mk = strtotime("\$end");

\$datediff = ( \$end_mk > \$begin_mk ? ( \$end_mk - \$begin_mk ) : ( \$begin_mk - \$end_mk ) );
\$days = ( ( \$datediff / 3600 ) / 24 );
\$days = \$days + 1; // to be inclusive of last date;
``````
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Question by:FrankTech
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Accepted Solution

FrankTech earned 0 total points
ID: 21824609
Well, I finally came up with this.  Anybody have a better or easier way?
``````<?php

// ** Set Beginning and Ending Dates, in YYYY-mm-dd format **
\$begin = '2008-01-01';
\$end = '2008-03-31';
\$begin_mk = strtotime("\$begin");
\$end_mk = strtotime("\$end");

// ** Calculate number of Calendar Days between the two dates
\$datediff = ( \$end_mk > \$begin_mk ? ( \$end_mk - \$begin_mk ) : ( \$begin_mk - \$end_mk ) );
\$days = ( ( \$datediff / 3600 ) / 24 );
\$days = \$days + 1; // to be inclusive of last date;

// ** Count days excluding Sundays **
\$iteration = 0;
\$numDaysExSunday = 0;
for (\$i=1; \$i<=\$days; \$i++) {
\$weekday = date("w", strtotime("\$begin + \$iteration day"));
echo "\$weekday<br>";
if (\$weekday !== '0') {
\$numDaysExSunday++;
}
\$iteration++;
}

// ** Output number of days excluding Sundays **
echo \$numDaysExSunday;
?>
``````
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