在一个laravel中连接两个不同的数据库5.5

Question

我想通过使用laravel 5.5中的简单代码删除if else语句来尽量减少这种情况。有人可以帮助我吗？

public function shirts($type='') 
{ 
    if($type == 'glass') { 
     $shirt = Product::where('category_id','1')->get(); 
     $products = Category::find(1); 
    } elseif ($type == 'ic') { 
     $shirt = Product::where('category_id','2')->get(); 
     $products = Category::find(2); 
    } elseif ($type == 'cover') { 
     $shirt = Product::where('category_id','3')->get(); 
     $products = Category::findOrFail(3); 
    } else { 
     $shirt = Product::all(); 
    } 
    return view('front.shirt',compact('products','shirt')); 
} 

Answer 1

一种方法是为您的类型创建映射并将类型与映射进行比较。

public function shirts($type = '') 
{ 
    $type_mappings = [ 
     'glass' => 1, 
     'ic' => 2, 
     'cover' => 3 
    ]; 

    if(array_key_exists($type, $type_mappings)) { 
     $shirt = Product::where('category_id', $type_mappings[$type])->get(); 
     $products = Category::find($type_mappings[$type]); 
    } else { 
     $shirt = Product::all(); 
     $products = null; 
    } 

    return view('front.shirt', compact('products', 'shirt')); 
} 

Answer 2

编辑：我假设你想避免if else没有什么问题的标题说，如果我仍不清楚，请添加评论，所以我可以更新的答案谢谢！

让我们处理它在其他地方，我猜你的函数只能有责任找到产品基于ID得到它不映射，因此，我们可以有这样的：

// Your function with single responsibility to return product. 
public function shirts($category = '') 
{ 
    $type = $this->CategoryIdMapper($category); 

    if($type == 0) { 
     $shirt = Product::all(); 
     $products = null; 
    } else{ 
     $shirt = Product::where('category_id',$type_id)->get(); 
     $products = Category::findOrFail($type_id); 
    } 
    return view('front.shirt',compact('products','shirt')); 
} 

//let the mapping part be done by some independent function which you can later modify accordingly and independently. 
public function CategoryIdMapper($category) 
{ 
    $categories = ['glass'=>1, 'ic'=> 2, 'cover' => 3 ]; 
    if(array_key_exist($category,$categories)) 
    { 
     return $categories[$category]; 
    } 
    return 0; 
}