10

I'm developing a pretty simple Symfony console application. It has just one command with one argument, and a few options.

I followed this guide to create an extension of the Application class.

This is the normal usage for the app, and it works fine:
php application <argument>

This also works fine (argument with options):
php application.php <argument> --some-option

If someone runs php application.php without any arguments or options, I want it to run as though the user had run php application.php --help.

I do have a working solution but it isn't optimal and is perhaps slightly brittle. In my extended Application class, I overrode the run() method as follows:

/**
 * Override parent method so that --help options is used when app is called with no arguments or options
 *
 * @param InputInterface|null $input
 * @param OutputInterface|null $output
 * @return int
 * @throws \Exception
 */
public function run(InputInterface $input = null, OutputInterface $output = null)
{
    if ($input === null) {
        if (count($_SERVER["argv"]) <= 1) {
            $args = array_merge($_SERVER["argv"], ["--help"]);
            $input = new ArgvInput($args);
        }
    }
    return parent::run($input, $output);
}

By default, Application::run() is called with a null InputInterface, so here I figured I could just check the raw value of the arguments and forcefully add a help option to pass to the parent method.

Is there a better way to achieve this?

gazareth
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  • i think your solution is just fine, why you think it´s brittle ? – john Smith Feb 04 '16 at 14:44
  • I don't like that it works outside of the existing input abstractions. It's also not the most readable code. If you see something like `if ($this->hasNoArguments()) { $this->addOption("help", true) }`, it's more readable. – gazareth Feb 04 '16 at 15:26

2 Answers2

11

I managed to work out a solution which didn't involve touching the Application class at all. To call the help command from within another command:

/**
 * @param InputInterface $input
 * @param OutputInterface $output
 * @return int
 * @throws \Symfony\Component\Console\Exception\ExceptionInterface
 */
protected function outputHelp(InputInterface $input, OutputInterface $output)
{
    $help = new HelpCommand();
    $help->setCommand($this);
    return $help->run($input, $output);
}
gazareth
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3

To do a specific action depending on command, you can use an EventListener which is called when the onConsoleCommand is fired.

The listener class should work as follows :

<?php

namespace AppBundle\EventListener;

use Symfony\Component\Console\Event\ConsoleCommandEvent;
use Symfony\Component\Console\Command\HelpCommand;

class ConsoleEventListener
{
    public function onConsoleCommand(ConsoleCommandEvent $event)
    {
        $application = $event->getCommand()->getApplication();
        $inputDefinition = $application->getDefinition();

        if ($inputDefinition->getArgumentCount() < 2) {
            $help = new HelpCommand();
            $help->setCommand($event->getCommand());

            return $help->run($event->getInput(), $event->getOutput());
        }
    }
}

The service declaration :

services:
     # ...
     app.console_event_listener:
         class: AppBundle\EventListener\ConsoleEventListener
         tags:
             - { name: kernel.event_listener, event: console.command, method: onConsoleCommand }
chalasr
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    Thanks. I think your answer, while more verbose and requiring an additional class, is likely more of the "right way" vs my own one below because it abstracts things a bit better. When I get the chance, I'll test it and if it's right, accept your answer. I'm away this week so it may be a while before I get to it :) – gazareth Feb 05 '16 at 15:00
  • You're welcome. My primary objective is that if people has the same kind of need, he can find an answer here. And now I know how use the built-in `HelpCommand`. So thank's for your good question – chalasr Feb 05 '16 at 15:08