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Start your free trialXing Li
2,112 PointsCreate a Fish object named $bass with the following data: name: "Largemouth Bass" flavor: "Excellent"
Create a Fish object named $bass with the following data: name: "Largemouth Bass" flavor: "Excellent" record: "22 pounds 5 ounces"
<?php
class Fish {
public $common_name = 'name';
public $flavor = 'flavor';
public $record_weight = 'weight';
function __construct($name, $flavor, $record){
$this->common_name = $name;
$this->flavor = $flavor;
$this->record_weight = $record;
}
}
?>
4 Answers
William Li
Courses Plus Student 26,868 PointsHi Xing Li
You may create a new object using the new statement, and pass along the required arguments to the constructor.
<?php
class Fish {
public $common_name = 'name';
public $flavor = 'flavor';
public $record_weight = 'weight';
function __construct($name, $flavor, $record){
$this->common_name = $name;
$this->flavor = $flavor;
$this->record_weight = $record;
}
}
// create a Fish object named $bass
$bass = new Fish("Largemouth Bass", "Excellent", "22 pounds 5 ounces");
?>
Xing Li
2,112 PointsTried it. It didn't work ;(
Xing Li
2,112 PointsGuess I forgot ";"
Victor Unda
1,367 PointsHere you go: Very simple
<?php
class Fish { public $common_name ="Largemouth Bass"; public $flavor = "Excellent"; public $record_weight = "22 pounds 5 ounces"; } // Creating an object
$bass = new Fish(); $bass->common_name; $bass->flavor; $bass->record_weight;
?>