PHP: If Result with Function

Question

I’m having a little problem using If with a Function to show a result.

function estaParaExpirar($data, $dias){
return(strtotime($data) < strtotime("+".$dias. "days") );
}


  if (estaParaExpirar($row[11], "10") ||
  estaParaExpirar($row[12], "10") ||  estaParaExpirar($row[13], "10") ||
  estaParaExpirar($row[14], "10") ||
  estaParaExpirar($row[15], "10") ||
  estaParaExpirar($row[16], "10") ||
  estaParaExpirar($row[17], "10")) {  
   $Nome1 = '<p>Nome: '.$row[10].'</p>' ;}

I did a little test with the code @Jader gave.

$row[11] = '0000-00-00';
$row[12] = '2014-08-15';
$row[13] = '2014-08-15';
$row[14] = '2014-08-15';
$row[15] = '2014-08-15';
$row[16] = '2014-08-15';
$row[17] = '2014-08-15';

if (($row[11] != '0000-00-00' and $row[12] != '0000-00-00' and $row[13] 
!= '0000-00-   00'      and $row[14] != '0000-00-00' and $row[15] != '0000-00-00'
and $row[16] != '0000-00-00' and $row[17] != '0000-00-00')
and ((estaParaExpirar($row[11], "10")) and (estaParaExpirar($row[12], "10")) and 
(estaParaExpirar($row[13], "10")) and
(estaParaExpirar($row[14], "10")) and   
(estaParaExpirar($row[15], "10")) and 
(estaParaExpirar($row[16], "10")) and
(estaParaExpirar($row[17], "10")))) { 

Upshot;

echo 'missing more than 10 days'; You are wrong

PS: I changed the code to but I am having trouble showing the result. If the result is 10 above the current date shows no result

     if (($row[11] != '0000-00-00' && estaParaExpirar($row[11], "10")) 
    || ($row[12] != '0000-00-00' && estaParaExpirar($row[12], "10"))
    || ($row[13] != '0000-00-00' && estaParaExpirar($row[13], "10")) 
    || ($row[14] != '0000-00-00' && estaParaExpirar($row[14], "10")) 
    || ($row[15] != '0000-00-00' && estaParaExpirar($row[15], "10")) 
    || ($row[16] != '0000-00-00' && estaParaExpirar($row[16], "10")) 
    || ($row[17] != '0000-00-00' && estaParaExpirar($row[17], "10"))) {     
    $Nome1 = '<p>Nome: '.$row[10].'</p>' ;}

Answer 1

Assuming I understand your problem, I have an alternative solution based on the class Datetime

Its problem seems to revolve around a comparison between the current date and a future date, ie a difference of intervals that, with the Datetime is resolved through Datetime::diff():

$currentDate = new DateTime;

$diff = $currentDate -> diff( new DateTime( $futureDate ) );

Simple as that! For a date 10 days in the future we will have in $diff:

DateInterval Object
(
    [y] => 0
    [m] => 0
    [d] => 9
    [h] => 7
    [i] => 14
    [s] => 29
    [weekday] => 0
    [weekday_behavior] => 0
    [first_last_day_of] => 0
    [invert] => 0
    [days] => 11
    [special_type] => 0
    [special_amount] => 0
    [have_weekday_relative] => 0
    [have_special_relative] => 0
)

The above output was generated based on a simple strtotime('+10 days'), hence the input d seems wrong to show 9, but that’s because the hours, minutes and seconds were not informed.

Its biggest problem, however, seems to be regarding the display. And, unfortunately, the Datetime It makes things a little difficult because there are no negative numbers for her. Instead, when the difference computed is prior to the (current) base date the input invert shall be changed from zero to one.

And that’s what you’d work your parole officers with:

if( $diff -> invert == 0 ) {

    if( $diff -> d > 10 ) {

        return sprintf( '%d days remaining', $diff -> d );

    } else {

        return sprintf( 'Will expire in %d days', $diff -> d );
    }

} else {

    return sprintf( 'Expired %d days ago', $diff -> d );
}

If the flag invert is zero, it means that the difference is positive, that is, the future date is really future. Hence it is a simple comparison: If the number of days is greater than the 10 you set, we show that X days to go. otherwise we show that the expires in X days.

Case to flag invert be 1, we have that the future date has passed, so we just show that expired x days ago.

As I do not know where this will be used, depending on the case you will not show that has already run out the deadline, getting this Else optional or for use other than in the form presented.

The complete code:

function showExpirationDate( $futureDate, $limit = 10 ) {

    $currentDate = new DateTime;

    $diff = $currentDate -> diff( new DateTime( $futureDate ) );

    if( $diff -> invert == 0 ) {

        if( $diff -> d > $limit ) {

            return sprintf( '%d days remaining', $diff -> d );

        } else {

            return sprintf( 'Will expire in %d days', $diff -> d );
        }

    } else {

        return sprintf( 'Expired %d days ago', $diff -> d );
    }
}

And the demonstration:

echo showExpirationDate( date( 'Y-m-d', strtotime( '+10 days' ) ) ); // Will expire in 9 days

Answer 2

Your function is not validating the input date, I added the check to return false if the date is invalid or null and added the default value of days:

function estaParaExpirar($data, $dias=10) {
    if (!strtotime($data) || empty($data)) return false;
    return(strtotime($data) < strtotime("+".$dias. "days") );
}

// Hoje é 12/08/2014

$row[11] = '0000-00-00';

if (estaParaExpirar($row[11])) {
    echo 'esta para expirar em menos de 10 dias';
} else {
    echo 'faltam mais de 10 dias';
}

//retorno
// faltam mais de 10 dias

$row[11] = '2014-08-20';

if (estaParaExpirar($row[11])) {
    echo 'esta para expirar em menos de 10 dias';
} else {
    echo 'faltam mais de 10 dias';
}

//retorno
// esta para expirar em menos de 10 dias

---

Bonus a function that returns the days to expire:

function diasParaExpirar($data) {
    if (!strtotime($data) || empty($data)) return false;
    return floor( ( strtotime($data) - strtotime(date('Y-m-d')) ) / 60 / 60 / 24 );
}

// Hoje é 12/08/2014

$row[11] = '2014-08-20';

echo 'faltam: '.diasParaExpirar($row[11]).' dias para expirar.';

//retorno
// faltam: 8 dias para expirar.