doumao6212 2015-05-05 08:22
浏览 328

php:退出php脚本后如何继续执行

I am writing a web page, which searches in a table and displays the data if all is well and a message if not. The code looks roughly like this:

<?php session_start(); ?>
<html><head> ..... </head>
    <body>
         <form action="search.php" method="post"> {parameters for searching} </form
         <?php include_once('display_results.php');?>
         <?php if (isset($_SESSION['message'])) {
                   echo "<script type=\"text/javascript\">alert(\"".$_SESSION['message']."\")</script>";
                   unset($_SESSION['message']);
               } ?>
     </body>
</html>

the display_results.php looks like this:

<?php
     //query table get results build output and if there was an error call do_exit($message)

 function do_exit($message) {
    //unset objects close connections and misc finalizations
    $_SESSION['message'] = $message;
    die();
 }

My problem is that when die() is called, it seems that not only does the display_results.php script stops, but also the original php file, so even though the message is set, there is no alert. On the other hand, if I remove the die() statement, and after each do_exit() I execute return this works correctly.

Now this isn't a huge problem, I can just make it so that I call return after each do_exit() but I was wandering if there was a way to exit the script but continue with the execution of the original php file

Thanks in advance for any help you can provide.

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3条回答 默认 最新

  • drl92080 2015-05-05 08:26
    关注

    why don't you use return false; at the end of your function do_exit()?

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