Calculate days between 2 dates and another criterial PHP MYSQL - php

Question

I'm new newbie to programming and I've got an error that i wasn't able to solve after hours searching.
I'm using a form to perform a search and display in a table the data, that part was fairly easy... but I also want to calculate how many days are in the final result, not in the dates I using for the first query.
The query searches for the result in days between 2 dates and from a specified company, the problem is in the final result it calculates how many days are from my result and not from the days i searched.
Example: If i search 2 dates between February 11 and February 20 Where the company = 1
If there is only 3 days result for Company 1, I want it to calculate only 3 Days.
I know my code is Wrong in that part but i just can't get it to do what i want.
Here is a little piece of my code to explain it:
if (isset($_POST['search'])) {
$date1 = mysql_prep($_POST['date1']);
$date2 = mysql_prep($_POST['date2']);
$latte = mysql_prep($_POST['latte']);
$query = "SELECT * FROM payroll WHERE (day BETWEEN '{$date1}' AND '{$date2}') AND (company = '{$latte}') ORDER BY day ASC ";
$result = mysql_query($query, $connection);
$woof = "SELECT SUM(hours) FROM (SELECT * FROM payroll WHERE (day BETWEEN '{$date1}' AND '{$date2}') AND (company = '{$latte}') ORDER BY day ASC) AS subt ";
$raw = mysql_query($woof, $connection);
if(!$raw) { die(mysql_error());}
$meow = mysql_result($raw, 0, 0);
$diff = abs(strtotime($date2) - strtotime($date1));
$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24) +1);
if(!$result) {
echo "FAIL";
} else {
echo "<table>
<tr>
<th> Date </th>
<th> Hours </th>
<th> Job Title </th>
<th> Job Description </th>
<th> Paid </th>
<th> For </th>
</tr>";
while($row = mysql_fetch_array($result))
{
$company = $row['company'];
if($company == 0) {
$company = "Wilson Electric";
} if($company == 1) {
$company = "Wilson Rental";
} if ($company == 2) {
$company = "Church of Christ";
}
echo "<tr>";
echo "<td class=\"center\">" . $row['day'] . "</td>";
echo "<td class=\"center\">" . $row['hours'] . "</td>";
echo "<td style=\"padding:5px;\">" . $row['job_title'] . "</td>";
echo "<td style=\"padding:5px;\">" . $row['job_description'] . "</td>";
echo "<td class=\"center\">" . $row['paid'] . "</td>";
echo "<td style=\"padding:5px;\">" . $company . "</td>";
echo "</tr>";
}
echo "<tr>";
echo "<td class=\"center\"> Total Days: " . $days . "</td>";
echo "<td class=\"center\"> Total Hours: " . $meow . "</td>";
echo "</tr>";
}
}
Thanks for Reading it, I hope someone can give me a solution, i know must be something really simple but I just couldn't figure out.

Answer 1

If you want a count of distinct day values that match the specification, then you could add a simple expression to the SELECT list of the "woof" SQL statement:
SELECT SUM(hours), COUNT(DISTINCT `day`) AS cnt_days FROM ...
$meow = mysql_result($raw, 0, 0);
$days = mysql_result($raw, 0, 1);
Note that if there is more than one row that has the same day value (which satisfies all of the predicates), that day value is counted only once. If you want a count of rows that have a non-NULL day value, including any duplicates, then omit the DISTINCT keyword.)
Note that the SQL statement could be more simply written; you could omit the (unnecessary) inline view (i.e. derived table) and the unnecessary ORDER BY ...
SELECT SUM(p.hours) AS sum_hours
, COUNT(DISTINCT p.day) AS cnt_days
FROM payroll p
WHERE p.day BETWEEN '{$date1}' AND '{$date2}'
AND p.company = '{$latte}'

Answer 2

May be you can use DateTime::diff