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I'm running ubuntu on WSL, and now I want to make my command to start up my php server(Nginx, mysql, php). Right now everytime I boot my pc I always run sudo service nginx start, sudo service mysql start, and the php.

I want to make my own custom command that when I run it, it will execute the starting services command. Until this point I know how to make custom command. But the problem is, I have 2 projects that need 2 different php version(One 7.1 and the other is 7.3). Can I make a command that accept arguments and run the service based on the arguments.

My expectation:

customcommand 7.1 will start nginx, mysql, and php7.1-fpm services

customcommand 7.3 will start nginx, mysql, and php7.3-fpm services

I know, I can just make different commands like customcommand7.1 and customcommand7.3. But if I can just make 1 command, why not?

Sorry if my question is a dump question. and Thanks! :D

  • in your case, just use a if / case statement (and don't forgot to start it using sudo) – damadam Dec 16 '19 at 15:06
  • @damadam how do you if / case in bash file? – Denny Rustandi Dec 16 '19 at 15:25
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Just save it in file some like function.sh and run it like: bash function.sh

#!/bin/bash

customcommand () {
    sudo service nginx start
    sudo service mysql start
    sudo service php$1-fpm start
    # ... use switch case
    echo "Run v$1"
}

# this command execute version 7.1
customcommand 7.1

if it work, put this in .bashrc if it linux, or .bash_profile in mac

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