Форум dkLab и Denwer

[faaliyet]

Hi there,
(I'm Sorry , if I post this message into wrong topic. I don't know russian language. )

-Check the line of JSHttpRequest.php file.

if (php_sapi_name() == "cgi") {

This function works until version PHP5.3.

Check this:
php.net/manual/en/function.php-sapi-name.php

Here is the fix: (+cgi, +cgi-fcgi)

$sapi_type = php_sapi_name();
if (substr($sapi_type, 0, 3) == 'cgi') {

[suhov]

this code from the jshttprequest example on your site
$JsHttpRequest =& new JsHttpRequest("windows-1251");

shall it be
$JsHttpRequest &= new JsHttpRequest("windows-1251");
?

[Maus]

suhov wrote:

shall it be $JsHttpRequest &= new JsHttpRequest("windows-1251");

No. It was PHP4 syntax ( "= new" produced object and its copy ). So, now syntax is:

Code (php):	скопировать код в буфер обмена
$JsHttpRequest = new JsHttpRequest("windows-1251");