Using loop variable outside the loop PHP

Question

This is probably very simple but I cannot figure it out and my searches have come up blank.

Here is what I have in mind:

<?php

$var1 = 0.0;
for ($i=1;$i<10;$i++){
    $var1 = $var1 + $i;
}
echo $var1[4]; // This would give 0, I believe.
?>

I hope my example made it clear what I'm trying to do, and I'm sure there is a simple solution, I just unfortunately cannot find it.

Thanks, Sam

Did you test it? Your example uses an array key [4] on a scalar variable $var1. If not for that, it would work. — Michael Berkowski
– Michael Berkowski, Commented Aug 7, 2012 at 20:04
As long as the variable is defined outside of the loop scope, it is accessible outside of the loop. — Matt
– Matt, Commented Aug 7, 2012 at 20:04
Sorry; I can't figure out what that's trying to do - you set up $var1 as an integer, but you're treating it as an array in the echo at the end? — andrewsi
– andrewsi, Commented Aug 7, 2012 at 20:05
@Matt in PHP, the scope isn't limited inside loops (like most other languages) - php.net/manual/en/language.variables.scope.php — newfurniturey
– newfurniturey, Commented Aug 7, 2012 at 20:06

Mihai Matei · Accepted Answer · 2012-08-07 20:14:39Z

8

<?php
  $arr = array();
  for ($i = 1; $i < 10; $i++){
      $arr[$i - 1] = $i;
  }
  echo $arr[4]; // This would give 5.
?>

edited Aug 7, 2012 at 20:14

answered Aug 7, 2012 at 20:06

Mihai Matei

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1 Comment

Sam Creamer Over a year ago

indeed, this is what I was going for though.

Micah Henning · Accepted Answer · 2012-08-07 20:07:14Z

0

You're not changing the value of $i in your loop, which would cause it to go on forever, I'm afraid. Also, you define $var1 as a number, but then you're trying to access it as though it were an array.

answered Aug 7, 2012 at 20:07

Micah Henning

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