Question

我收到PHP通知错误。这段代码在php 5.3中运行良好，但随后我将PHP升级到PHP 7.我想要做的是，从链接中获取URL，然后只显示随URL附带的参数。这是代码。

的index.php

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

bootstrap.php中

<?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";
    }
?>

转到网址时的输出：localhost / myDir / index.php / abc / def

注意：未定义的索引：第8行的/srv/http/myDir/bootstrap.php中的控制器
注意：未定义的索引：第14行的/srv/http/myDir/bootstrap.php中的操作

主页
索引

Answer 1

测试empty() ... true为0，'0'，false，''，空数组（） ......通知也消失了！ ...为你的其他ifs和数组索引做同样的事情！

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

为了避免类似的警告，您还应该在方法，函数等中提供默认值：

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

Answer 2

如果您的代码运行正常＆amp;问题只是删除通知错误然后你可以在php脚本中使用error_reporting(0)。

添加error_reporting(0)作为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";
    }
?>

如何解决这个PHP通知错误？

3 个答案: