遍历php中类的每个实例

Question

如何根据给定的代码遍历类的每个实例：

class Project {
    // Constructor
    public function __construct($projectTitle, $projectSummary, $projectTools) {
        $this->projectTitle = $projectTitle;
        $this->projectSummary = $projectSummary;
        $this->projectTools = $projectTools;
    }

我创建了以下公共函数：

public function displayProjectCard() {
        echo '<div class="card mt-4 w-75" >
                <div class="card-body">
                    <h5 class="card-title">' . $this->projectTitle . '</h5>
                    <p class="card-text">' . $this->projectSummary . '</p>
                    <p class="tools">' . $this->projectTools . '</p>
                </div>
            </div>';
    }

...但是我不确定如何-或是否有可能-遍历每个类实例，以便每个“ projectCard”都显示在其自己的Bootstrap卡中。

实例的创建方式如下：

$project1 = new Project(
            "Project Title", 
            "This is where you'll write a description of your project and related features.", 
            "Tools", 
        );

目标是使结果看起来像这样：

This image

Answer 1

根据您的代码，我将采用以下方式：

// instantiate projects array like this
$projects = array(
    new Project("Title1", "Summary", "Tools"),
    new Project("Title2", "Summary", "Tools"),
    new Project("Title3", "Summary", "Tools"),
    new Project("Title3", "Summary", "Tools"),
);

// then iterate each $projects and then call the displayProjectCard()
foreach ($projects as $project) {
    $project->displayProjectCard();
}