Re: Include a B.php but not influence A.php

From: Date: Thu, 19 May 2016 06:43:16 +0000
Subject: Re: Include a B.php but not influence A.php
References: 1 2  Groups: php.general 
Request: Send a blank email to php-general+get-325601@lists.php.net to get a copy of this message
On May 18, 2016, at 10:46 PM, Jigar Dhulla wrote:
On Thu, May 19, 2016 at 6:56 AM, 酷丁 <zsxx51@vip.qq.com> wrote:
hi,everyone! a problem has troubled me for a long time,I have searched ,but haven't found the answer. some thing like this: ---A.php <?php include 'b.php'; echo 'test A.php';//because calling exit in b.php,so this can't be execute ?> ---B.php <?php exit("test B.php"); ?>
can you comment out portions of script for testing? Try and find the code in B.php that actually calls exit and comment it out. If you don't need to include B.php at the top, put it after echo 'test A.php'; //If code in A.php doesn't depend on B.php Other wise it looks like you have situation that won't work regardless of what you do with it. JK
I want to include b.php(b.php is a third library,but it will exit the script) but no die.I try to find some function that can execute b.php and not die a.php ,but I can't. are there any other method can do this? thk...
​Hi, Libraries generally never die, they only have classes to include in project. According to me, it must be a full-fledged application/script, so what I can suggest is that you run the script externally. Means, if your app is located at http://<your-domain>.com then that script can be http://<script>.<your-domain>.com or http://<your-domain>.com/<script>. After setting up like this you can make curl or ajax request to 3rd party script URL to fetch the output. I personally do it that way. Hope this helps.​ --Regards, Jigar Dhulla


« previous php.general (#325601) next »