Question

php里面存放对象，不能调用用对象的函数

<?php

require 'MyHomework1/myfunction/db_class.php';

interface Member
{

function getID();
function getName();
function getLevel();

}

class Student implements Member
{
private $ID,$name,$level;

public function __construct($ID,$name,$level)
{
$this->ID = $ID;
$this->name = $name;
$this->level = $level;
//echo "student";

}

public function getID() { return $ID; }

public function getName() { return $name; }

public function getLevel() { return $level; }

public function myTeamMate()
{
$id = $this->ID;
$team_mate = array();

$team_table = new SubTable("TEAM");

$team = mysql_fetch_array( $team_table->query("WHERE MemberID = '$id'") );
$team_name = $team['TeamName'];
$team = $team_table->query("WHERE TeamName = '$team_name'");

while ($row = mysql_fetch_array($team))
{
$student = new Student($row['MemberID'],$row['MemberName'],1);
array_push($team_mate,$student);

}
return $team_mate;

}

}

class TA implements Member
{
private $ID,$name,$level;

public function __construct($ID,$name,$level)
{
$this->ID = $ID;
$this->name = $name;
$this->level = $level;
//echo "ta";

}

public function getID() { return $ID; }

public function getName() { return $name; }

public function getLevel() { return $level; }

}

class MemberFactory
{
public static function factory($ID,$name,$level){

if ($level == 1)
return new Student($ID,$name,$level);

else return new TA($ID,$name,$level);

}

}

$student = MemberFactory::factory("50","lujunjie","1");
$a = $student->myTeamMate();

echo sizeof($a);

foreach ($a as $aa){
echo $aa->getID();

}

foreach ($a as $aa){
print_r($aa);

}

?>

我把好多个Student类放到$a里面，这句不执行echo $aa->getID();其他打印出来没错，为什么？

Answer 1

public function getID() { return $ID; }
修改为：
public function getID() { return $this->ID; }