Calculation of hours for point PHP + MySQL
I am creating a table for calculating employee hours worked, 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 is like this:
On the site will be displayed this way:
In the last column, in "hours worked" I need to put the result of the calculation of Input 1 (beginning of the day), output 1 (Lunch), Input 2 (Return of lunch) and output 2 (end of Day)
I used this select to grab all the data from the table:
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 Am 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 calculate this?
0
Author: Denis Rudnei de Souza, 2017-12-27
1 answers
Opa, what you want is the result in hours from these strings. It can 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;
0
Author: David Alves, 2017-12-27 19:40:27