如何解决此 PHP 通知错误？

Question

我收到 PHP 通知错误。这段代码在 php 5.3 中运行良好，但后来我将 PHP 升级到 PHP 7。我想做的是，从 link，只显示URL附带的参数。这是代码。

index.php

<?php 
    require_once('bootstrap.php');
    $bootstrap = new Bootstrap($_GET);
?>

bootstrap.php

<?php 
class Bootstrap{
    private $controller;
    private $action;
    private $request;
    public function __construct($request){
        $this->request = $request;
        if($this->request['controller'] == ''){
            $this->controller = "Home";
        }
        elseif($_GET($request['controller'])){
            $this->controller = $this->request['controller'];
        }
        if($this->request['action'] == ''){
            $this->action = "index";
        } else{
            $this->action = $this->request['action'];
        }
        echo "<br />$this->controller<br />$this->action";
    }
?>

转到 URL 时的输出：localhost/myDir/index.php/abc/def

注意：未定义索引：controller in /srv/http/myDir/bootstrap.php on line 8
注意：未定义索引：第 14

行 /srv/http/myDir/bootstrap.php 中的操作

首页
指数

Answer 1

测试 empty() ... 将 true 用于 0、'0'、false、''、空数组() ...而且通知也不见了！ ...对您的其他 ifs 和数组索引执行相同的操作！

if(empty($this->request['action'])) {

为了避免类似的警告，您还应该在您的方法、函数等中提供一个默认值。:

function ($arg=FALSE, $arg2=TRUE, $arg3=5, ...) {

Answer 2

如果您的代码工作正常并且问题只是为了消除通知错误，那么您可以在 php 脚本中使用 error_reporting(0)。

将 error_reporting(0) 添加为 php 脚本中的第一个语句

Answer 3

测试是否设置： isset($this->request['action']) isset($this->request['controller'])

像这样：

<?php 
class Bootstrap{
    private $controller;
    private $action;
    private $request;
    public function __construct($request){
        $this->request = $request;
        foreach ($request as $key => $value) {
            echo $key . " = " . $value;
        }
        if(isset($this->request['controller']) && $this->request['controller'] == ''){
            $this->controller = "Home";
        }
        elseif(isset($this->request['controller']) && $_GET($request['controller'])){
            $this->controller = $this->request['controller'];
        }
        if(isset($this->request['action']) && $this->request['action'] == ''){
            $this->action = "index";
        }
        else{
            $this->action = $this->request['action'];
        }
        echo "<br />$this->controller<br />$this->action";
    }
?>

如何解决此 PHP 通知错误？

How to solve this PHP notice error?

php

compiler-errors

get

notice