Determine Apache version on Windows?

From CMD:

"C:\Program Files\[Apache directories]\bin\httpd.exe" -v

Example:

C:\Documents and Settings\User>"c:\Program Files\Apache Software Foundation\Apache2.2\bin\httpd.exe" -v
Server version: Apache/2.2.21 (Win32)
Server built:   Sep  9 2011 10:26:10

The version will be published in the log specified under ErrorLog within your configuration. (/var/log/httpd/error_log is a standard location in many Linuxes.)

Additionally, you should be able to execute the httpd binary equivalent in windows with -v.

Example from Linux:

# /usr/local/apache/bin/httpd -v
Server version: Apache/1.3.41 (Unix)
Server built:   Sep 10 2009 10:42:36

I wrote this as an alternative to the suggestions above and for those using XAMPP.

Navigate to your XAMPP folder, enter your Apache folder, enter your bin folder, right click on httpd.exe. The tabs on the top will be General/ Compatibility/ Details/ Previous Versions, we want the Details tab and there it will list the File version and Product version. They will say the same thing with the File version having an extra decimal place.

For Apache itself, it will be the same procedure excluding the XAMPP part.

If you can't recall or can't find your Apache or XAMPP folder just do a windows search for httpd.exe and right click and chose the open file location.

Enable server-status

The server-status page will also show this, in addition to performance information. It starts with something like this:

Apache Server Status for localhost

Server Version: Apache/2.2.14 (Win32) PHP/5.3.1

Server Built: Sep 28 2009 22:41:08

See Detailed Instructions for server-status on another question.

You can use this command in powershell ise

(((. "C:\Apache24\bin\httpd.exe" -v ) -split "/") -split " \(")[1]

With Powershell you don't even need to run the .exe, you can directly read the version attribute.

(Get-Command ./httpd.exe).FileVersionInfo.FileVersion