Start an apache server in any directory from command line

Question

I want to be able to start an apache server from the command line, typing something like apache site-folder or apache . --port=2000

This should read and use .htaccess files.

I know about python -m SimpleHTTPServer and it's close to what I need, but not quite.

Ideal solutions:

1. Contributing a great command line interface to apache itself
2. Writing a simple command line tool that wraps/contains apache (or something)
3. Linking to docs on an existing cli for apache

I just want to type command <Directory> --port=8000 --other-options

The command name could also be pache

At some point I may want to use this in production. It should be easy to send the process to the background, and then stop that instance or all instances, like forever

Relevant links: http://httpd.apache.org/docs/2.4/programs/httpd.html

Also

It should be only one command for anyone to install the script for immediate use

Answer 1

What about apache debug mode (-X) ?

apache2 -X -d. -f.htaccess -C"PidFile `mktemp`"  -C"Listen 1025" 
-C"ErrorLog /dev/stdout" -C"DocumentRoot `pwd`"

to put it in the background once started you may use Ctrl^Z then type "bg"

Using the the FOREGROUND flag, wrapping it up in a shell script:

#!/bin/sh
if [ $# -ne 2 ]; then 
    echo "$0 <port> <dir>"
    exit 10 
fi
/usr/sbin/apache2 -DFOREGROUND -d. -f.htaccess -C"PidFile `mktemp`" \ 
-C"Listen $1" -C"ErrorLog /dev/stdout" -C"DocumentRoot $2" -e debug

call it "pache", chmod +x, then you can run

./pache 1026 /tmp/webroot