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Start your free trialDevin Scheu
66,191 PointsPhp help
Question: Create a method on Fish named getInfo that takes no parameters and returns a string that includes the common_name, flavor, and record_weight for the fish. When called on $bass, getInfo might return "A Largemouth Bass is an Excellent flavored fish. The world record weight is 22 pounds 5 ounces."
Code:
<?php
class Fish {
public $common_name = "";
public $flavor = "";
public $record_weight = 0;
function __construct($name, $flavor, $record) {
$this->common_name = $name;
$this->flavor = $flavor;
$this->record_weight = $record;
}
public getInfo() {
return "A {$common_name} is an {$flavor} fish. The world record weight is {$record_weight}";
}
}
$bass = new Fish("Largemouth Bass", "Excellent", "22 pounds 5 ounces");
?>
4 Answers
David Tonge
Courses Plus Student 45,640 PointsFigured it out for you. I would say becareful when learning multiple languages at once. I have the same problem with syntax.
<?php
public function getInfo() {
return "A". $this->common_name . " is an " . $this->flavor .
" flavored fish. The world record weight is " . $this->record_weight;
}
?>
David Tonge
Courses Plus Student 45,640 PointsYou're missing the word "flavored" after the $flavored variable.
edit: Ryan Duchene answered the question
Devin Scheu
66,191 PointsStill not working for me
Ryan Duchene
Courses Plus Student 46,022 PointsYou forgot to add "flavored" in your return string, right after {$flavor}
. :)
<?php
// ...
return "A {$common_name} is an {$flavor} flavored fish. The world record weight is {$record_weight}";
// ...
Ryan Duchene
Courses Plus Student 46,022 PointsRyan Duchene
Courses Plus Student 46,022 PointsAh. Of course. The
$this->
part is what does it.Good job, that one had me stumped.