Loop autoincrementing date

Question

How can I make a auto increment date per month and showing the last day of the month

Example:

2017-01-31

2017-02-28

2017-03-31

$parcelas  = 12;
$data_inicial = 2017-01-31
for ($i = 1; $i <= $parcelas; $i++){        
    // abaixo o auto increment
    echo $data_inicial.........??????
}

Answer 1

To always catch the last date of the month use date with mktime():

date('t', mktime());

Generate plots by the last day of the month according to the information of the year:

function gerar_parcelas($y = 2017) 
{
    $parc = array();        
    for($i = 1; $i < 13;$i++){
        $parc[] = date('Y-m-t', mktime(0,0,0,$i,1,(int)$y));
    }
    return $parc;
}

var_dump(gerar_parcelas(2017));

Example

---

In this other code is generated as if it were instalments of payment, that is to say, whether 13/12/2005 and the amount of plots the algorithm generate from that date only changing the month and year if it goes to the next. This code also fits your doubt if you are informed the last date of the month with 12 plots.

Generate plots by a date:

<?php

function parcelas($data, $numero = 12)
{
    $parc = array();
    $parc[] = $data;
    list($ano, $mes, $dia) = explode("-", $data);
    for($i = 1; $i < $numero;$i++)
    {
        $mes++;
        if ((int)$mes == 13)
        {
            $ano++;
            $mes = 1;
        }
        $tira = $dia;
        while (!checkdate($mes, $tira, $ano))
        {
            $tira--;
        }
        $parc[] = sprintf("%02d-%02d-%02d", $ano, $mes, $tira);
    }
    return $parc;
}


$data = "2017-01-31";

var_dump(parcelas($data, 12));

Example

^{In this second example is generated by the date so to work the same is in the question always has to pass the last day of the first month of the year, because, this code generates for any day of the month other than the first that takes the year and generates the last date of the month of that year.}

References:

• date
• mktime
• checkdate

Answer 2

You can do it like this

<?php
$parcelas = 12;
$data_inicial = '2017-01-31';
for ($i = 1; $i <= $parcelas; $i++){        
    echo $data_inicial."<br>";
    $d = new DateTime( $data_inicial );
    $d->modify( 'last day of next month' );
    $data_inicial = $d->format( 'Y-m-d' );
}

Answer 3

Kill a fly with a cannonball?! Well, I thought of a very simple way to solve your problem that would be you use strtotime(). See the code below:

<?php    
function returnDates($parcelas, $ultima_data){
    for ($i = 1; $i <= $parcelas; $i++){   
       $date = strtotime("+$i month", strtotime($ultima_data));
       echo date("Y-m-t", $date)."\n";
    } 
}

echo returnDates(20,"2007-02-12");

In the above example I call the function by placing the number of installments and the initial date.

See the example functioning on the ideone.

Answer 4

         $m=0; 
         for($i=1;$i <= $nparcelas; $i++){
            $a1 = new DateTime($data_inicial);
            $a2 = new DateInterval('P'.$m.'M');
            $a1->add($a2);
            $a1->format("Y-m-d") -> formatação datetime para inserir no banco
            $m++;
         }