Time calculation for PHP + MySQL point

0

I am creating a table for calculating employee hours, but I am not able to calculate to see if the correct amount of hours worked has been worked on the day.

The database looks like this:

Thesitewilllooklikethis:

In the last column, under "Working Hours", you need to enter the calculation result for Input 1 (Start of day), Output 1 (Lunch), Input 2 (Return from Lunch) and Output 2 (end of day) p>

I used this select to get all the data in 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'm displaying 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?

    
asked by anonymous 27.12.2017 / 15:52

1 answer

0

Oops, 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;
    
27.12.2017 / 17:11