Calculation of hours for PHP + MySQL point
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I am creating a table for calculating working hours of employees, but I am not able to do the calculation to know if the correct amount of hours was worked in the day.
In the database it is like this:
The site will be displayed this way:
In the last column, in "Worked Hours" I need to put the result of the calculation of Entry 1 (Start of the day), Exit 1 (Mouse), Entry 2 (Return of the Lunch) and Exit 2 (end of the day)
I used this select to get all the table data:
SELECT p.id_ponto, p.tb_user_id_user, p.tb_ponto_mes_id_ponto_mes, p.dia_ponto, p.dt_entrada1, p.dt_saida1, p.dt_entrada2, p.dt_saida2, u.id_user, u.nm_user FROM tb_ponto p INNER JOIN tb_user u on u.id_user = p.tb_user_id_user
I'm displaying the values like this:
echo '<tr><td>' . $item["nm_user"] . '</td>' . '<td>' . $item["dia_ponto"] . '/' . $item["tb_ponto_mes_id_ponto_mes"] . '/2017' . '</td>' . '<td>' . $item["dt_entrada1"] . '</td>' . '<td>' . $item["dt_saida1"] . '</td>' . '<td>' . $item["dt_entrada2"] . '</td>' . '<td>' . $item["dt_saida2"] . '</td>' . '<td>8 Horas</td></tr>';
How can I do that?
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Opa, what you want is the result in hours from these strings. It is possible to be done as follows:
// Faz o cálculo das horas $total = (strtotime($item['dt_saida1']) - strtotime($item['dt_entrada1'])) + (strtotime($item['dt_saida2']) - strtotime($item['dt_entrada2']));
// Encontra as horas trabalhadas
$hours = floor($total / 60 / 60);// Encontra os minutos trabalhados
$minutes = round(($total - ($hours * 60 * 60)) / 60);// Formata a hora e minuto para ficar no formato de 2 números, exemplo 00
$hours = str_pad($hours, 2, "0", STR_PAD_LEFT);
$minutes = str_pad($minutes, 2, "0", STR_PAD_LEFT);// Exibe no formato "hora:minuto"
echo $hours.':'.$minutes;