PHP loop to get todays date and the following 9 days

Question

I have the following php code:

$date = new DateTime();
$date->setTimeZone(new DateTimeZone('America/New_York'));
$index = 0;


while( $index <10){
    $date->modify('+'. $index .' day');
    print $date->format('Y-m-d') . ' - index: ' . $index . '<br/>';
    $index++; 
}

What this script should do: it should display todays date and the next 9 dates for the following days: ( I also printed the $index value each time to make sure the index it's right all time).

What it actually displays:

2017-05-29 - index: 0
2017-05-30 - index: 1
2017-06-01 - index: 2
2017-06-04 - index: 3
2017-06-08 - index: 4
2017-06-13 - index: 5
2017-06-19 - index: 6
2017-06-26 - index: 7
2017-07-04 - index: 8
2017-07-13 - index: 9

NOTE: See how from date 2017-06-01 jumps to 2017-06-04 and then to 2017-06-08 and so on.

I have no idea why this happens, can someone tell me? Is this even the right approach ?

Answer 1

As you want to add 1 day on every iteration, you don't need to use the index on the sum, just use $date->modify("+1 day"); .

Also, put the line where you print the date before you modify it.

What happens in your code is that you start adding 0 on the first iteration, which then you print the actual day. Next iteration, you add 1, so tomorrow is printed. Next iteration, you add 2, and skip 1 day. Next, add 3, and skip 2 days... And this over and over.

Answer 2

不要更改$index的值使其等于1，甚至不要尝试在modify()函数中执行“ +1天”

Answer 3

for($i=1;$i<=10;$i++)
echo date("y-m-d",strtotime($i." day"))."<br />";

Answer 4

Use this loop

$begin = new DateTime("2017-05-29");
$end = new DateTime("2017-06-8");
$end = $end->modify('+1 day');
$interval = DateInterval::createFromDateString('1 day');

$index = 0;
foreach ($period as $dt) {
    if ($index == 9) {
        break;
    }

    echo $dt->format("Y-m-d");
    $index = $index + 1;
}