方法名必须是字符串,PHP中的错误

Question

这是我的代码,

class Tasks {
    public $parent;

    public function __construct($parent) {
        $this->parent = $parent;
    }

    public function get_task($parent) {
        return mysqli_query($db, "SELECT task, status, created_at FROM tasks WHERE parent IS '$parent' AND user='$user_id'");
    }
}

$project = new Tasks("NULL");

print "<div class='project'>" . $project->$get_task() . "</div>";

我希望这应该将"NULL"传递给mysql_query并返回结果.但我得到错误,

Fatal error: Method name must be a string

更新:为了解决这个问题,我只是没有创建类的实例.相反,我将我想要的值传递给函数Tasks::get_task(value-here).

Answer 1

更换

print "<div class='project'>" . $project->$get_task() . "</div>";

同

print "<div class='project'>" . $project->get_task() . "</div>";

(删除$符号).