如何在laravel中優化查詢？

Question

您好我希望儘量減少這樣的代碼：

if (empty(Input::get('channelName'))) { 
    $channels = Channel::orderBy('updated_at', 'desc')->paginate($perPage)->setPath('?limit=' . $perPage); 
} else { 
    $channels = Channel::where('name', 'like', '%' . Input::get('channelName') . '%')->orderBy('updated_at', 'desc')->paginate($perPage)->setPath('?limit=' . $perPage); 
} 

Answer 1

$channels = Channel::orderBy('updated_at', 'desc')->paginate($perPage)->setPath('?limit=' . $perPage); 

    if (empty(Input::get('channelName'))) { 
     $channels = $channels->get(); 
    } else { 
     $channels = $channels->where('name', 'like', '%' . Input::get('channelName') . '%')->get(); 
    } 

Answer 2

的Query Builder是一個很酷的「事情」，因爲它可以讓你你需要鏈的方法。這使得根據我們的需要靈活地追加查詢語句。

$channelsQuery = Channel::orderBy('updated_at', 'desc'); 
if (Input::has('channelName')) { 
    $channelsQuery = $channelsQuery->where('name', 'like', '%' . Input::get('channelName') . '%'); 
} 
$channels = $channelsQuery->paginate($perPage)->setPath('?limit=' . $perPage); 

你如果還可以使用Input::has('channelName')而不只是empty(Input::get('channelName'))。