Php calculete the cost for each day passed

Question

  $1date =$row['Date1'];
     $2date = $row['Date2'];
       $datediff = $1date - $2date;
echo $datediff;

I want to count the days between and put into a table the result(10 dollars for each day passed)

If you're working with mysql, then do it directly in sql, not in php ... — Stefan Brendle, Apr 24 '13 at 20:45

score 1 · Answer 1 · answered Apr 24 '13 at 20:50

1

You could do it in MySQL then work it into a variable. Do the initial call

SELECT TIMESTAMPDIFF('Date1','Date2');

answered Apr 24 '13 at 20:50

Kevin Lynch

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score 1 · Accepted Answer · answered Apr 24 '13 at 20:50

1

// convert to unix timestamp
$1date = strtotime($row['Date1']);
$2date = strtotime($row['Date2']);

// 86400 seconds in a day
// floor to round down, change to ceil to round up
$datediff = floor(($1date - $2date) / 86400);

$cost = $days * 10;

answered Apr 24 '13 at 20:50

Samutz

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jx12345 · Answer 3 · 2013-04-24T20:52:47.143

0

Try:

$days = date_diff(date_create($row['Date1']), date_create($row['Date2']))->format('%a');
$cost = $days * 10;

edited Apr 24 '13 at 20:52

answered Apr 24 '13 at 20:45

jx12345

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i get an error at the first line – marina Apr 24 '13 at 20:49
whats the error message? – jx12345 Apr 24 '13 at 20:49
Sorry i missed off a closing bracket before the ->format – jx12345 Apr 24 '13 at 20:52
i've edited to include it, could you try it again? – jx12345 Apr 24 '13 at 20:53

Php calculete the cost for each day passed

3 Answers3