Question

我试图写一个工匠命令。我已经掌握了基础知识。

现在我试图通过将一些代码移动到另一个文件来清理它。

问题是，在其他文件中，像$this-line('hello')这样的命令不起作用。

有没有一种简单的方法可以让它发挥作用？

（下面有两个文件，第一个文件是命令，它可以工作。请注意“工作”的底部。文件，我们做 $tmp = new viewclass然后$tmp->display()

第二个文件是我要放置所有逻辑的地方 - 如何调用继承的函数，如$ this-＆gt; line，$ this-＆gt; info，$ this-＆gt; table等...来自那个第二档？

http://laravel.com/docs/5.1/artisan#writing-output

CrudFromDb_view.php:

<?php
namespace path\laravel_crudfromdb\Commands;

use Illuminate\Console\Command;
use \path\laravel_crudfromdb\Classes\viewclass;

class CrudFromDb_Views extends Command
{
    protected $signature = 'z:viewviews';
    protected $description = 'Displays generated views on screen. Does not change or create any files';

    public function __construct()
    {
        parent::__construct();
    }

    public function handle()
    {
        $this->line(' THIS LINE WORKS');

        $tmp = new viewclass;
        $tmp->display();  // <- Fails, see 'viewclass.php' file below
    }
}

viewclass.php:

<?php
namespace path\laravel_crudfromdb\Classes;

class viewclass extends Command
{

    protected $env;
    protected $dbhost;
    protected $dbname;
    protected $dbuser;
    protected $dbpw;
    protected $connection;

    function __construct()
    {    
      //  parent::__construct();
    }

    function display()
    {
            //this fails!
            //how can I call this 'line' function?
            $this->line('This is a line');
    }
}

注意，一种有效的方法是传递对象。

（感觉PHP应该有内置的方法来实现这一点吗？）

即CrudFromDb_view.php：

$tmp = new viewclass($this);
$tmp->display()

并在viewclass.php

class viewclass
{
    protected $myparent;
    function __construct($thisfromparent)
    {
        $this->myparent = $thisfromparent
    }

    function display()
    {
       $this->myparent->line('this works');
    }
}

是否有更优雅的方式来做上面做过的事情？我已尝试过parent :: line（＆＃39; text＆＃39;）但这不起作用： - （

Answer 1

在viewclass.php中添加以下命名空间，并允许viewclass扩展Command。

use Illuminate\Console\Command;
class viewclass extends Command
{
    protected $env;
    protected $dbhost;
    protected $dbname;
    protected $dbuser;
    protected $dbpw;
    protected $connection;

    function __construct()
    {    
        //parent::__construct();
    }

    function display()
    {
        //this fails!
        //how can I call this 'line' function?
        $this->line('This is a line');
    }
}

Answer 2

这个怎么样？

<?php
namespace path\laravel_crudfromdb\Commands;

use Illuminate\Console\Command;
use \path\laravel_crudfromdb\Classes\viewclass;

class CrudFromDb_Views extends Command
{
    protected $signature = 'z:viewviews';
    protected $description = 'Displays generated views on screen. Does not change or create any files';

    public function handle()
    {
        $this->line('THIS LINE WORKS');

        $tmp = app()->make(viewclass::class);
        $tmp->display();  // Should work
    }
}

使用app()->make()将确保Laravel实现父Command类中所需的所有依赖注入。这是Laravel的核心目的，它首先是IoC（控制反转）容器。

要使其更加“犹太洁食”，请将您的viewclass类添加为依赖注入...

<?php
namespace path\laravel_crudfromdb\Commands;

use Illuminate\Console\Command;
use \path\laravel_crudfromdb\Classes\viewclass;

class CrudFromDb_Views extends Command
{
    protected $signature = 'z:viewviews';
    protected $description = 'Displays generated views on screen. Does not change or create any files';

    protected $viewclass;

    public function __construct(viewclass $viewclass)
    {
        $this->viewclass = $viewclass;
        parent::__construct(); //put it as the last line to avoid constructor bugs.
    }
    public function handle()
    {
        $this->line('THIS LINE WORKS');
        $this->viewclass->display();  // Should still work thanks to IoC Container
    }
}

Laravel PHP工匠命令 - 尝试使用来自被调用类的输出命令

注意，一种有效的方法是传递对象。

2 个答案: