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Start your free trialMathew Coulter
1,998 Pointscreate a fish object named $bass with the following data name: "Largemouth Bass" flavor: "Excellent" record: "22 pounds
I don't know why it isn't accepting my code. Any help?
<?php
class Fish {
public $common_name;
public $flavor;
public $record_weight;
function __construct($name, $flavor, $record){
$this->common_name = $name;
$this->flavor = $flavor;
$this->record_weight = $record;
}
$bass = new Fish("Largemouth Bass","Excellent","22 pounds 5 ounces");
}
?>
1 Answer
Mathew Coulter
1,998 PointsI figured it out. I forgot that $bass had to be outside of the class Fish.