我试图制作一个表,但对于不同的页面,我需要不同类型的变量。我想通过在父类中进行总体布局来完成此操作,然后将所有细节放在扩展类中。但是,我不知道如何从扩展类中获取信息返回到父类。如何从扩展类中获取变量到其父类中
父类:
class table {
protected $tablename;
protected $mysqli;
function set_name($name) {
$this->tablename = $name;
}
function connectdb($mysqli) {
$this->mysqli = $mysqli;
}
function make_table($tablename) {
//all table making stuff
//here I want to access the completed variable
}
扩展类:
class tasktable extends table {
public $completed;
function set_completed($completed) {
$this->completed = $completed;
echo $completed;
}
function get_completed() {
return $this->completed;
}
}
代码的网页上:
$tasktable1 = new tasktable($tableName);
$tasktable1->connectdb($mysqli);
$tasktable1->set_completed(0);
$tasktable1->make_table($tableName);
嗯......'$ this-> whatever'? – 2014-10-02 14:03:40
逻辑上,必须在父类中访问的属性应在父类中定义,但是当前的代码将工作得很好 - 请参阅:http://codepad.viper-7.com/UIHgFc – Steve 2014-10-02 14:15:25