erlang - startup script

Question

To start my program I do the next sequence:

$ erl
> c(module1).
> c(module2).
> c(modulen).
modulen:start().

Is there any possibility to create script that allow me to launch my program ?

Answer 1

You could use a loader script that takes care of starting your application in an OTP fashion:

-module(XYZ_app).

-export([start/0]).

start() ->
    application:start(inets),
    application:start(XYZ).

You launch that script through a shell script. Beware of using escript if you plan on building daemons running at the OS boot time as they are tricky.

#!/bin/bash
erl -boot start_sasl -s XYZ_app start

Of course you need your XYZ.app file (just an example):

{application, tinycouch,
 [{description, "tinycouch"},
  {vsn, "0.1"},
  {modules, [
         tinycouch, tinycouch_app, tinycouch_sup,
         tinycouch_server, tinycouch_logger_h, tinycouch_utils,
         tinycouch_db, mod_tinycouch
        ]},
  {registered, [tinycouch
            ,tinycouch_server
            ,tinycouch_sup
           ]},
  {applications, [kernel, stdlib, sasl, inets]},
  {env, []},

  %% Application Start point
  {mod, {tinycouch_sup, []}}]}.

... and all your .erl files must have been compiled.

Note that if you intend to distribute your application (e.g. Debian repository etc), you probably should consider having a make file to compile and install your app.

As a final note: you could also go with a boot file (the ultimate OTP way) but I find those very constraining: a boot file ties your application to specific version release of Erlang (and other dependent applications). Your application might be able to run in various Erlang releases but you will need to have a separate build/release process for each "platform release" you intend to ship for.

Answer 2

You can compile the modules with erlc

erlc module1 module2 module3

You can also create a script to run you program

#!/usr/bin/escript

main(_) ->
    modulen:start().

Then in the console:

chmod +x start.erl
./start.erl