Average days from dates Mysql, PHP

Question

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Calculate DateDiff in SQL in Days:Hours:Mins:Seconds format

I have this MySql query:

SELECT DATEDIFF(MAX(hit_date), MIN(hit_date)) / (COUNT(hit_date) - 1) AS hitavg FROM my_table

This returns a value (average days between recorded rows from hit_date column, this column is in TIMESTAMP format, so YY:MM:DD HH:MM:SS)

Assuming that this value is 385.500 (returned from the query), how can I format in PHP this number as 385 Days, "n" Hours (where "n" is the decimal value, in this case 500)?

Thanks in advance to all!

Do you simply want days and hours, or did you want days, hours, minutes, seconds? — crush, Jan 23 '13 at 21:34

crush · Accepted Answer · 2013-01-23T21:41:57.467

2

Well, if you only want Days and Hours, then:

$value = 385.500;

$days = (int) $value;
$hours = 24 * ($value - $days);

echo "$days Days, $hours Hours";

edited Jan 23 '13 at 21:41

answered Jan 23 '13 at 21:35

crush

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I replaced `floor()` with `(int)` because casting to int is 2x faster than calling `floor()` and achieves the same result. – crush Jan 23 '13 at 21:42

Average days from dates Mysql, PHP

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