Re: Calling servlet in PHP
| From: | jeremy brand | Date: | Fri, 15 Dec 2000 22:05:05 +0000 |
| Subject: | Re: Calling servlet in PHP | ||
| References: | 1 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-30595@lists.php.net to get a copy of this message | ||
Hi Joseph,
It looks like should wrap your encrypted data with urlencode(). Some
of thge bytes may not be valid bytes for the URL.
-jeremy brand
Jeremy Brand :: Sr. Software Engineer :: 408-245-9058 :: jeremy@nirvani.net
http://www.JeremyBrand.com/Jeremy/Brand/Jeremy_Brand.html
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On Fri, 15 Dec 2000, Paulson, Joseph V. "Jay" wrote:
> Date: Fri, 15 Dec 2000 15:59:34 -0600
> From: "Paulson, Joseph V. \"Jay\"" <jay.paulson@viatel.com>
> To: php-general@lists.php.net
> Subject: [PHP] Calling servlet in PHP
>
> Hello everyone--
> We seem to be having some trouble calling a java servlet from a PHP
> application. The PHP application is encrypting the account number and the
> division and then passing it in the URL. The java servlet then decrypts the
> information and serves up a page. This however is not happening and a blank
> page is being served. We believe it has something to do with the class paths
> used by PHP and Java. Have any of you had experience working with PHP in
> connection with Java or have an idea on what the trouble may be?
>
> Here's the code that we are refering to when we call the servlet. The
> strange thing is without the encryption of $ACCTNO and $DIV and pass it in
> the URL it works fine.
>
> Code Snippett:
> $ACCTNO = encrypt($userInfo["acctNum"]);
> $DIV = encrypt($userInfo["div"]);
> $PAGE_TYPE = "Payments";
> header("Location:/servlet/CSWPayments?ACCTNO=$ACCTNO&DIV=$DIV&PAGE_TYPE=$PAG
> E_TYPE");
>
> Thanks for any help,
> Jay Paulson
>
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