将一个Var从Class传递给Include

Question

我有一个函数，我想使用一个包含内，但我需要所有的变量是公共的类被干净地传递到页面布局包含文件。

变量在函数内部传递，而不是在传递给include时。有什么简单的建议吗?

private function printGraph() {
    /*
    if($_SERVER['HTTP_HOST']=='localhost') {
        echo "<pre>n";
        echo "Actual: ".$this->actual."n";
        echo "Actual Bar: ".$this->actualbar."n";
        echo "Attainable Bar: ".$this->attainablebar."n";
        echo "Attainable: ".$this->attainable."n";
        echo "ADSL2Calc: ".$this->adsl2calc."n";
        echo "</pre>";
    }
    */
    //Adding the new look Tshooterlayout!
    include 'tshooterlayout14.php';
    // Actual Speed info and bar
    $actual=file_get_contents("inc/troubleshooter/tshoot3-actual.inc");
    $actual=sprintf($actual,$this->actual,$this->actualbar,$this->attainablebar);

Answer 1

直接使用$this:

test.php

<?php
class Test {
    public $foo = 'bar';
    public function testme(){
        include "include.php";
    }
}
$T = new Test();
$T->testme();

include.php

<?php
echo "You got your " . $this->foo ." in my foo!n";

结果

$ php test.php
You got your bar in my foo!