PHP exercise loops

Question

Using while loop to print 10-20 odd numbers on the screen, with <-> (exapmle 11-13-15-17-19, <-> not after 19).How to take the last number so as not to put a -. Thank you.

<?php

$x = 10;
while($x < 20){
   $x++;
   if($x % 2){
      echo $x. "-";
   }
}

Tip: Print the `-` _before_ $x...unless it's the first one - which is much easier to detect than the last one. — ADyson, Oct 26 '22 at 18:41
Try to use `break` after that condition, if you want the code to exit, that will stop the loop. You should dig into a php starting guide/lesson for such questions. Here a link for the reference: https://www.php.net/manual/en/control-structures.break.php — Gkiokan, Oct 26 '22 at 18:42
You can [push them to an array](https://www.php.net/manual/en/function.array-push.php) and then use [implode()](https://www.php.net/manual/en/function.implode.php) with the `-` as glue. Just make sure you uncomment the code first (remove the `//`). Weird to have them in the posted code at all, tbh. — M. Eriksson, Oct 26 '22 at 18:44

Max Pattern · Accepted Answer · 2022-10-26T22:06:57.300

1

You can push the odd values to an array and after the loop you can convert the array to a string with the implode (https://www.php.net/manual/en/function.implode.php) function.

$x = 10;
$arr = [];
while($x<20){
  $x++;
  if ($x%2){
    $arr[] = $x;
  }
}
echo implode(",", $arr);
// output will be a comma seperated string

without (helper)array You can use rtrim() (https://www.php.net/manual/de/function.rtrim.php) funktion.

$str = "1,2,3,";
echo rtrim($str,",");

edited Oct 26 '22 at 22:06

answered Oct 26 '22 at 18:48

Max Pattern

1,430
7
18

Thank you very much. But how to solve without array? – Andrey Oct 26 '22 at 18:50
– Andrey Oct 26 '22 at 18:50
@Андрей You welcome. I updated my answer with a solution without array. – Max Pattern Oct 26 '22 at 22:07

score 1 · Answer 2 · answered Oct 26 '22 at 18:54

As mentioned in the comments, have a boolean variable say firstTime. If this is true, don't prepend a hyphen, else if it is false, prepend it with a hyphen.

<?php

$x = 10;
$firstTime = true;
while($x < 20){
   $x++;
   if($x % 2){
      echo ($firstTime ? "" : "-") . $x;
      $firstTime = false;
   }
}

score 0 · Answer 3 · answered Oct 26 '22 at 19:03

Turn the limit number into a variable and use a ternary operator to print out a dash only if $x + 1 < $limit

https://paiza.io/projects/EDj6-u-FAcYxYoR7ON_cvg

<?php

$x = 10;
$y = 20;
while($x < $y){
   $x++;
   if($x % 2){
      echo ($x + 1 === $y) ? $x : $x. "-";
   }
}

?>

score 0 · Answer 4 · answered Oct 26 '22 at 19:42

Simple approach

Imagine you are printing a list of numbers from 10 to 20. If it's an even number, print the "-" symbol and if it's odd number, print the number.

<?php
$start = 10;
$end = 20;
while($start<$end)
{
  $n=$start;
  if($n>10)
  echo ($n%2==0)?"-":$n; //the gist is here
  $start++;
}

PHP exercise loops

4 Answers4

Simple approach