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Start your free trialCliff Jackson
2,887 PointsDon't understand how to output $checker?
Either my concatenation is wrong or i am not understanding the question any ideas??
<?php
include "class.palprimechecker.php";
$checker = new PalprimeChecker;
$checker->number = 17;
echo "The number " . $checker . ";
echo "(is|is not)";
echo " a palprime.";
?>
2 Answers
jamesjones21
9,260 Pointsyou will need to reference the following to get the number that being set when you are setting $checker->number
echo "The number " . $checker->number;
Cliff Jackson
2,887 PointsThanks James