PHP 仅从 function 中退出当前包含的脚本文件

Question

I know I can use return to exit current script file, but what if we want to exit current script from inside a function?

For example, we define a exitScript() function that prints closing html tags and exits current script file, but if we use return it only exits from current function.

Answer 1

You can use Exceptions to do this, not particularly elegant, but this should do what your after, replace these methods.

public function fn1()
{
    try {
        $fn2 = $this->fn2();
    }
    catch ( Exception $e )  {
    }

    echo 'I want this to be displayed no matter what!';
}


public function fn4()
{
    $random = rand(1, 100);

    if ($random > 50)
    {
        return true;
    }
    else
    {
        // I want to exit/break the scirpt to continue running after
        // the $fn2 = $this->fn2() call in the $this->fn1() function.
        //exit();
        throw new Exception();

        echo "This shouldn't be displayed.";
    }
}

Answer 2

You can return from an included file to the 'main file', skipping execution of the remaing part after the return .

So this will output: ABXend , skipping C

a.php

<?php 
  echo 'A';
  include('b.php');
  echo 'end';
?>

and b.php

<?php 
   echo 'B';
   if(!x(6))return; //return here

   echo 'C';

   function x($num){
     echo 'X';
     if($num!==1)return false;
     }
?>