Question

我使用此代码时出错

致命错误：第8行/home/u448966927/public_html/Myclass.php中只能通过引用传递变量

Myclass.php

<?php

class yt
{
    public $channels = array
    (
        "music" => array( "id" => 74,"img" => "<img src=../icon/music.png height=18 width=18 />", "name" => "Music", "slug" => "music") ,
        "sports" => array( "id" => 11,"img" => "<img src=../icon/sport.png height=18 width=18 />", "name" => "Sports", "sug" => "sport"),
        "movies" => array( "id" => 45,"img" => "<img src=../icon/movies.png height=18 width=18 />", "name" => "Movies", "sug" => "movie"),
        "gaming" => array( "id" => 85,"img" => "<img src=../icon/game.png height=18 width=18 />", "name" => "Gaming", "sug" => "game"),
        "people" => array( "id" => 56,"img" => "<img src=../icon/people.png height=18 width=18 />", "name" => "People & Blog", "sug" => "people"),
        "comedy" => array( "id" => 72,"img" => "<img src=../icon/comedy.png height=18 width=18 />", "name" => "Comedy", "sug" => "comedy"),
        "news" => array( "id" => 84,"img" => "<img src=../icon/news.png height=18 width=18 />", "name" => "News & Politics", "sug" => "news"),
        "animation" => array( "id" => 29,"img" => "<img src=../icon/film.png height=18 width=18 />", "name" => "Film & Animation", "sug" => "animation"),
        "auto-vehicles" => array( "id" => 59,"img" => "<img src=../icon/auto.png height=18 width=18 />", "name" => "Autos & Vehicles", "sug" => "auto-vehicles"),
        "howto" => array( "id" => 68,"img" => "<img src=../icon/howto.png height=18 width=18 />", "name" => "Howto & Style", "sug" => "howto"),
        "science" => array( "id" => 98,"img" => "<img src=../icon/science.png height=18 width=18 />", "name" => "Science & Technology", "sug" => "science")

    );
    public function getChannels(){
        $html = "<li><a href='#' class='current'><img src=../icon/pupolar.png height=18 width=18 />\tPopular</a>";
        ksort(channels);
        foreach (channels as $channel) {
            $html .="<li><a href='channel/".$channel["slug"]."'>now</a>";
        }
        return $html;
    }
};

?>]

的index.php

<?php include_once('Myclass.php'); echo getChannels(); ?>

Answer 1

从

更改此段代码

&lt;?php include_once(&apos;ytcx.php&apos;); echo getChannels(); ?&gt;

要

<?php 
    include_once('ytcx.php'); 
    $obj = new yt();
    echo $obj->getChannels();
?>

我认为PHP认为'是传递的参数，&是PHP中的引用字符，它只是让人感到困惑。

您还需要更改

public function getChannels(){
        $html = "<li><a href='#' class='current'><img src=../icon/pupolar.png height=18 width=18 />\tPopular</a>";
        ksort($this -> channels);
        foreach ($this -> channels as $channel) {
            $html .="<li><a href='channel/".$channel["slug"]."'>now</a>";
        }
        return $html;
    }

当您从类方法中引用类属性时，您需要$this->。

Answer 2

channels在您的ksort和foreach（第22行）中不存在且无效

更改

ksort(channels);
foreach (channels as $channel) {

要

ksort($this->channels);
foreach ($this->channels as $channel) {

由于函数getChannels在一个类中，您需要instantiate the class。在index.php中添加以下内容;

$objYT = new yt;
echo $objYT->getChannels();

https://eval.in/415737

Answer 3

试试这个..

class yt
{
    public $channels = array
    (
        "music" => array( "id" => 74,"img" => "<img src=../icon/music.png height=18 width=18 />", "name" => "Music", "slug" => "music") ,
        "sports" => array( "id" => 11,"img" => "<img src=../icon/sport.png height=18 width=18 />", "name" => "Sports", "slug" => "sport"),
        "movies" => array( "id" => 45,"img" => "<img src=../icon/movies.png height=18 width=18 />", "name" => "Movies", "slug" => "movie"),
        "gaming" => array( "id" => 85,"img" => "<img src=../icon/game.png height=18 width=18 />", "name" => "Gaming", "slug" => "game"),
        "people" => array( "id" => 56,"img" => "<img src=../icon/people.png height=18 width=18 />", "name" => "People & Blog", "slug" => "people"),
        "comedy" => array( "id" => 72,"img" => "<img src=../icon/comedy.png height=18 width=18 />", "name" => "Comedy", "slug" => "comedy"),
        "news" => array( "id" => 84,"img" => "<img src=../icon/news.png height=18 width=18 />", "name" => "News & Politics", "slug" => "news"),
        "animation" => array( "id" => 29,"img" => "<img src=../icon/film.png height=18 width=18 />", "name" => "Film & Animation", "slug" => "animation"),
        "auto-vehicles" => array( "id" => 59,"img" => "<img src=../icon/auto.png height=18 width=18 />", "name" => "Autos & Vehicles", "slug" => "auto-vehicles"),
        "howto" => array( "id" => 68,"img" => "<img src=../icon/howto.png height=18 width=18 />", "name" => "Howto & Style", "slug" => "howto"),
        "science" => array( "id" => 98,"img" => "<img src=../icon/science.png height=18 width=18 />", "name" => "Science & Technology", "slug" => "science")

    );

    public function getChannels(){
        $html = "<li><a href='#' class='current'><img src=../icon/pupolar.png height=18 width=18 />\tPopular</a>";
        ksort($this->channels);
        foreach ($this->channels as $channel) {
            $html .="<li><a href='channel/".$channel["slug"]."'>now</a>";
        }
        return $html;
    }
}

-----关于index.php ---

  include_once('Myclass.php'); 
    $yt = new yt;    
    echo $yt->getChannels();

Answer 4

您发布的代码中存在多个问题。

ksort(channels);
foreach (channels as $channel) {

PHP中的变量名称以$开头。当PHP达到channels时，它认为它是使用define()创建的常量。但是函数`ksort()需要通过引用传递的类型数组的参数。这是错误的来源。

如果将其更改为ksort($channels)，则仍然是错误，因为当前范围中没有定义变量$channels。你可能需要的是使用当前对象的$channels属性，并且总是使用$this（当前对象），然后是->（成员访问运算符）和您要访问的媒体资源的名称，不包含$（channels）：

ksort($this->channels);

下一行也一样：

foreach ($this->channels as $channel);

index.php：

中也存在类似的问题

<?php include_once('Myclass.php'); echo getChannels();

对getChannels()的调用尝试执行全局函数getChannels()但它失败，因为不存在这样的函数。您要完成的是调用类getChannels()的对象的方法yt。为此，您必须首先创建这样的对象：

<?php
include_once 'Myclass.php';
$obj = new yt();
echo $obj->getChannels();
// That's all

文件Myclass.php应如下所示：

<?php

class yt
{
    // ...

    public function getChannels(){
        $html = "<li><a href='#' class='current'><img src=../icon/pupolar.png height=18 width=18 />\tPopular</a>";
        ksort($this->channels);
        foreach ($this->channels as $channel) {
            $html .="<li><a href='channel/".$channel["slug"]."'>now</a>";
        }
        return $html;
    }
}

数组输出不能从类php工作

4 个答案: