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jonny black
279 PointsIt keeps saying I haven't changed $integerOne. so confused
I am following the previous video to do this task yet i seem to be doing something wrong. $integerOne = 1; $integerTwo = 2;
var_dump( $integerOne + 5); var_dump( $integerTwo - 1);
<?php
//Place your code below this comment
$integerOne = 1; ($integerOne + 5);
$integerTwo = 2; ($integerTwo - 1);
$floatOne = 1.5;
?>
1 Answer
Jennifer Nordell
Treehouse TeacherHi there, jonny black ! You added 5 to $integerOne but then nothing happens with the results. You didn't echo or var_dump them so we can't see the results and they are not saved back into $integerOne.
You could do this:
$integerOne += 5;
Or this:
$integerOne = $integerOne + 5;
These two lines do the same thing, They add 5 onto $integerOne and then assign that result back into the $integerOne value overwriting the original. You will need this same type of thing for $integerTwo.
Hope this helps!
