如何在另一个类中调用另一个类的函数?
这是我目前的代码:
<?php
class database
{
private $host = "localhost";
private $user = "root";
private $password = "";
private $db = "9dot";
//connect to database
public function connect()
{
$connect = mysql_connect($this->host,$this->user,$this->password);
$database = mysql_select_db($this->db,$connect);
}
//Get exam headers
public function GetExam()
{
$this->connect();
$select = "SELECT e.ExamHeaderID, CONCAT(e.Firstname,' ',e.Lastname) AS Fullname e.Age, e.Position, e.Date FROM examheader e";
$result = mysql_query($select);
$table = "";
$table .= "<th>ExamID</th>";
$table .= "<th>Name</th>";
$table .= "<th>Age</th>";
$table .= "<th>Position</th>";
$table .= "<th>Date</th>";
$exam_id = "";
$fullname = "";
$age = "";
$position = "";
$date = "";
while($row = mysql_fetch_array($select))
{
$exam_id = mysql_real_escape_string($row['ExamHeaderID']);
$fullname = mysql_real_escape_string($row['Fullname']);
$age = mysql_real_escape_string($row['Age']);
$position = mysql_real_escape_string($row['Position']);
$date = mysql_real_escape_string($row['Date']);
}
}
public function GetExamDetails($ExamID)
{
$this->connect();
//$select = "SELECT ed.ExamDetailsID, ed.";
}
//Save Exam header
public function SaveExam($firstname,$lastname, $age, $position)
{
$this->connect();
$date = date("m/d/Y");
mysql_real_escape_string($firstname);
mysql_real_escape_string($lastname);
mysql_real_escape_string($age);
mysql_real_escape_string($position);
$insert = "INSERT INTO examheader (Firstname,Lastname, Age, Position, Date) VALUES ('$firstname','$lastname','$age','$position')";
$result = mysql_query($insert);
$exam_id = mysql_insert_id();
//return exam id for insertion of exam details
return $exam_id;
}
public function SaveExamAnswer()
{
}
}
?>
<?php
include 'database.php';
$database = new $database();
class department
{
$database->//property or function
}
?>
如何从数据库类
调用函数答案 0 :(得分:5)
您可以在部门类中声明数据库实例,如下所示:
class department
{
public function __construct()
{
$database = new database();
$database->someFunction();
}
}
如果你不想直接在你的部门类中使用它,你可以使用global
系统声明它,假设包含了所需的文件,如下所示:
include 'database.php'
$database = new database();
class department
{
public function __construct()
{
global $database;
$database->someFunction();
}
}
答案 1 :(得分:3)
你可以扩展它们。
class department extends database {
function test() {
$this->getdbfunction();
}
}
或者你通过构造函数给他们:
class department {
private $db;
public function __construct(Database db) {
$this->db = $db
}
}
$department = new department($database);
答案 2 :(得分:2)
您可以定义属性,该属性是第一个类的实例,然后将方法应用于第二个类示例中的已定义属性:
class A {
function a () {
}
...
}
class B {
$attClassA;
function __construct(A objA) {
$this->attClassA = objA;
}
// you can use the attClassA in any class you want and then apply the chosen method
...
}
答案 3 :(得分:1)
尝试这样的事情(粗略的例子):
<?php
include 'database.php';
class department
{
public function doSomething() {
$database = new $database();
$val = $database->GetExam();
//do something
}
}
?>
答案 4 :(得分:1)
创建一个'database'类型的对象,然后使用它调用该方法:
$database = new database();
$connection = $database->connect();
答案 5 :(得分:0)
您可以执行以下操作:
如果您不想手动包含PHP文件,请使用spl_autoload_register()函数 否则,您需要包含具有要从其中导入方法的类的PHP文件。
file-A.php-包含类A的PHP文件
<?php
class A{
function DoSquare($x){
return $x*$x;
}
}
?>
file-B.php-包含类B的PHP文件
假设您正在使用spl_autoload_register():
<?php
class B{
private $ObjectA = NULL;
public function __construct(){
$this->ObjectA = new A();
}
public function PrintSquare($x){
echo $this->ObjectA->DoSquare($x);
}
}
?>
如果手动包含PHP文件:
<?php
require("file-A.php");
class B{
private $ObjectA = NULL;
public function __construct(){
$this->ObjectA = new A();
}
public function PrintSquare($x){
echo $this->ObjectA->DoSquare($x);
}
}
?>