Add 1 month function in php fails when current month has no day 31

Question

In this code PHP, I am checking the type of recurrence of an activity contained in a list of activities, and adding 1 month when it is monthly recurrence, 2 months when it is bimonthly recurrence and etc.

if($recorrencia == "UNICA")
        return;
    elseif($recorrencia == "MENSALMENTE")
        $data_vencimento = date('Y-m-d H:m:s', strtotime('+1 month', strtotime($atividade['DT_VENCIMENTO']))); /*1 mês*/
    elseif($recorrencia == "BIMESTRALMENTE")
        $data_vencimento = date('Y-m-d H:m:s', strtotime('+2 month', strtotime($atividade['DT_VENCIMENTO']))); /*2 mês*/
    elseif($recorrencia == "TRIMESTRALMENTE")
        $data_vencimento = date('Y-m-d H:m:s', strtotime('+3 month', strtotime($atividade['DT_VENCIMENTO']))); /*3 mês*/
    elseif($recorrencia == "SEMESTRALMENTE")
        $data_vencimento = date('Y-m-d H:m:s', strtotime('+6 month', strtotime($atividade['DT_VENCIMENTO']))); /*6 mês*/    
    elseif($recorrencia == "ANUALMENTE")
        $data_vencimento = date('Y-m-d H:m:s', strtotime('+12 month', strtotime($atividade['DT_VENCIMENTO']))); /*12 mês*/

However, the PHP looks like it’s actually adding 30 days, because when I have dates like the following:

31/05/2019

After rotating the function, the date becomes:

01/07/2019

I understand that is why in the month 06 does not have day 31, because all other days add correctly.

How could I get around this problem?

Answer 1

I have the following suggestion, which I believe solves your problem clearly.

if($recorrencia == "UNICA") return;

$recorrencias           = [
    'MENSALMENTE'       => 1,
    'BIMESTRALMENTE'    => 2,
    'TRIMESTRALMENTE'   => 3,
    'SEMESTRALMENTE'    => 6,
    'ANUALMENTE'        => 12,
];
$meses                  = $recorrencias[ $recorrencia ];
$data_vencimento        = new \DateTime( $atividade['DT_VENCIMENTO'] );
$ano                    = $data_vencimento->format( 'Y' );
$mes                    = $data_vencimento->format( 'm' );
$dia                    = $data_vencimento->format( 'd' );

// Adiciona a quantidade de meses pretendida, mas define o dia como o último do mês
$data_vencimento->setDate($ano, $mes + $meses + 1, 0);

// Se o dia pretendido for menor que o último dia do mês, então faz a correção
if ( $dia < $data_vencimento->format( 'd' ) ) $data_vencimento->setDate( $ano, $mes + $meses, $dia );

$data_vencimento        = $data_vencimento->format( 'Y-m-d H:i:s' );

Answer 2

First, the question is: what "adding 1 month". 1 month = 30 days? 28 days? 29 days? 31?

The following code:

echo date("Y-m-d", strtotime("2019-01-31 +1 month"));

It has the following result:

2019-03-03

It is correct because the function converts the second part of the date (the month) to the next), which would be 02/31/2019. As there are 31 days in February and this year the month had 28 days, are added the missing days, playing the date for the day 3 March (if you do the same test with 2016, you will see that the final date is 2 March because February had 29 days in 2016).

Therefore, I think you should reformulate your question and/or need.

• If you want to add 30 (or 60 or 90 days) to your date, simply add the number of days according to your need.
• If you simply want to say that "bimonthly" is today + 2 months, add 2 to the current month.

There are other approaches too, but they all depend on the answer to the question "added 1 month" to contextualize correctly.