Question

控制器：

public function leaderboard() {
    $users = DB::table('users')
        ->join('images', 'users.id', '=', 'images.user_id')
        ->select(DB::raw('users.*, COUNT(*) AS totalUploads'))
        ->orderby('totalUploads', 'desc')
        ->groupby('users.id')
        ->take(50)
        ->get();

    return View::make('leaderboard')->with('users', $users);
}

查看：

    <tbody>
        <?php $i = 0; $rank = 0; $lastCount = null; ?>
        @foreach($users as $user)
        <?php $i++; if ($user->totalUploads != $lastCount) { $rank = $i; } else { $rank = null; } ?>
        <tr>
            <td>{{ $rank }}</td>
            <td><a href="user/{{ $user->username }}">{{ $user->username }}</a></td>
            <td>{{ $user->totalUploads }}</td>
        </tr>
        <?php $lastCount = $user->totalUploads; ?>
        @endforeach
    </tbody>

当前输出： （破折号表示<tr>＆＃39> s

Rank  Name  Uploads
------------------- 
   1  Joe       103
-------------------   
   2  Dan        99
-------------------   
   3  Pam        87
-------------------   
      Kim        87
-------------------  
      Bob        87
-------------------      
   6  Mel        60

目标输出： （破折号表示<tr>＆＃39> s

Rank  Name  Uploads
------------------- 
   1  Joe       103
-------------------   
   2  Dan        99
-------------------   
   3  Pam        87
      Kim        87
      Bob        87
-------------------      
   6  Mel        60

基本上我想添加一个foreach或其他类型的循环，如果<br />被绑定，则会在<tr> $user->totalUploads内将用户与{{1}}分开。

Answer 1

编辑：

<tbody>
        <?php $i = 0; $rank = 0; $lastCount = null; ?>
        @for($i=0; $i<count($users); $i++)

        <?php $user = $users[$i]; 
            if ($user->totalUploads != $lastCount) { $rank = $i; } else { $rank = null; } 
        ?>

         <tr>
            <td>{{ $rank }}</td>
            <td>
                <a href="user/{{ $user->username }}">{{ $user->username }}</a>

                <?php $j=0; ?>

                @if(isset($user[$i+1]) && $users[$i+1]->totalUploads == $user->totalUploads)

                    @for($j=$i+1; $j<count($users)&&$users[$j]->totalUploads == $user->totalUploads; $j++)

                        <a href="user/{{ $users[$j]->username }}">{{ $users[$j]->username }}</a>


                    @endfor

                @endif

            </td>
            <td>{{ $user->totalUploads }}

                @if($j>0)

                    @for($k=0; $k<$j-$i; $k++)

                        {{ $user->totalUploads }}


                    @endfor

                @endif

            </td>
        </tr>

        <?php 
        $i=$i+$j;
        $lastCount = $user->totalUploads; ?>
        @endfor
    </tbody>

请注意我没有测试过这段代码，即使这看起来很复杂，我认为它会起作用

OLD VERSION

您可以尝试以下内容：

 <tbody>
        <?php $i = 0; $rank = 0; $lastCount = null; ?>
        @foreach($users as $user)
        <?php $i++; if ($user->totalUploads != $lastCount) { $rank = $i; } else { $rank = null; } ?>
        @if($rank) <tr> @else <br/> @endif
            <td>{{ $rank }}</td>
            <td><a href="user/{{ $user->username }}">{{ $user->username }}</a></td>
            <td>{{ $user->totalUploads }}</td>
        @if($rank)</tr> @endif
        <?php $lastCount = $user->totalUploads; ?>
        @endforeach
    </tbody>

如果在foreach中有条件，基本上会添加一个刀片。如果等级没有改变（意思是它在同一＆＃34;等级＆＃34;，你不输出但是输出一个
。结束标记只会是当rank不为null（不在同一级别）时应用

在foreach循环中有一个foreach循环

1 个答案: