Solution! RE: [ciapug] Compatibility[?] issue: apache2.2.2/php4.4.2/winxp

[Henke, Ken]

I'm a longtime lurker, and now a contributor. ;-)
I was having the exact same problem you were having, I found the
solution to the problem running apache2.2.2/php4.4.2/winxp
 
In a nutshell, make sure you have the exact php4apache2.dll compiled for
the version of Apache (the version with the php zip is not correct for
2.2.2), and if you don't have .NET installed, you have to get the VC++
redistributable package.
The answer and how-to is in the forum and thread listed below.
The forum refers to Apache 2.2.0, but the downloads page does have a
2.2.2 version of php4apache2.dll
 
http://www.apachelounge.com/forum/viewtopic.php?t=113
and
http://www.apachelounge.com/download/

________________________________

From: ciapug-bounces at cialug.org [mailto:ciapug-bounces at cialug.org] On
Behalf Of Jane Swanson
Sent: Tuesday, May 30, 2006 12:30 PM
To: ciapug at cialug.org
Subject: Re: [ciapug] Compatibility[?] issue: apache2.2.2/php4.4.2/winxp


Thanks for your help, but I was not successful with this combo.  Couldnt
find abachectl, but found:  " In general, httpd should not be invoked
directly, but rather should be invoked via apachectl
<http://httpd.apache.org/docs/2.2/programs/apachectl.html>  on
Unix-based systems or as a service on Windows NT, 2000 and XP
<http://httpd.apache.org/docs/2.2/platform/windows.html#winsvc>  and as
a console application on Windows 9x and ME
<http://httpd.apache.org/docs/2.2/platform/windows.html#wincons> . "[
http://httpd.apache.org/docs/2.2/programs/httpd.html  ] The "service"
is MS Services.  Can start/stop apache there or from the Apache Service
Monitor -- but only if [LoadModule php4_module "c:/php/php4apache2.dll"]
is not there.  
 
Don't know how to tell wich MPM service is running or how to direct
prefork to run.  Don't know why I care to run the latest & "greatest".
 
Gave up on apache2.2.2.  Installed 1.3.34.  Works fine.
 
Again, thanks for your attention.  I continue to watch this forum for
ideas.
 
Jane Swanson
 


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