TinyButStrong創建動態<th>與動態<tr>和​​循環問題

Question

我想創建員工在日曆表視圖中作爲附加圖像的離開報告。

我有來自控制器取出僱員，並存儲在和陣列。
同樣的方式我提取日期（基於選定的年份和月份）在控制器中並存儲在一個數組中。
基於此員工和日期，我正在獲取每個員工的葉子並將其存儲在數組中。

控制器

$month = 7; 
    $year = 2015; 
      for($d=1; $d<=31; $d++)//for dynamic dates as displayed in image 
      { 
       $time=mktime(12, 0, 0, $month, $d, $year);   
       if (date('m', $time)==$month)  
       $this->data['blk4'][]['date']=date('d', $time); 
       $this->data['blk5'][]['day']=date('D', $time); 
       $this->data['blk6'][]['fulldates']=date('Y-m-d',$time); 
      } 
      $data['employee'] = $this->teamprofile_model->allTeamMembers('team_profile_full_name', 'ASC');//fetching all amployee 

      foreach($data['employee'] as $d)//for each employee checking the leave 
      { 
       foreach($this->data['blk6'] as $date)//for each date 
       { 
        $this->db->where('employee_id',$d['team_profile_id']); 
        $this->db->where('leave_date',$date['dates']); 
        $Q=$this->db->get('leave_reports'); 
        if($Q->num_rows() > 0) 
        { 
         foreach ($Q->result_array() as $row) 
         { 
          $data1[] = $row; 
         } 
        } 
        $this->data['blk8'][]['leave'] = $data1[0]['leave_time'];//stored leave of each employee in this array 
        $data1=""; 
       } 
      } 

HTML

<table class="footable table table-bordered table-hover" border="1"> 
     <thead> 
      <tr> 
       <th>Employee</th>//display employee 
       <th><!--[blk4.date;block=th;comm]--><br/><!--[blk5.day;block=th;comm]--></th> //display dates as displying in above image 
      </tr> 
     </thead> 
     <tbody class=""> 
      <tr> 
       <td><!--[blk7.all_team_members;block=tr;comm]--></td>//dynamic employee list 
       <td>here i want to display the leave stored in [blk8.leave]</td>      
      </tr> 
     </tbody> 
</table> 

但問題是，[blk8.leave]存儲所有員工離開每個日期的，如果我打印爲[blk8 .leave; block = td; comm]然後它將所有數組值打印在一行中。我想打破這陣一個月的結束日期是31

輸出應該是：

Answer 1

的問題是當它需要你離開數據線性結構與日期關聯。

TBS（TinyButStrong）作爲內置功能，用於將表與動態列（或其他類似結構）合併。

下面的例子與你的問題非常相似，你可以很容易地適應它。 http://www.tinybutstrong.com/examples.php?e=dyncol1

但是您的數據結構應該修改。 這裏是一個數據如何可能的例子：

 $blk7 = array(); 
     foreach($data['employee'] as $d)//for each employee checking the leave 
     { 
      $employee = array(
       'team_profile_id' => $d['team_profile_id'], 
       'all_team_members' => $d['all_team_members'], 
      ); 
      foreach($this->data['blk6'] as $date)//for each date 
      { 
       $this->db->where('employee_id',$d['team_profile_id']); 
       $this->db->where('leave_date',$date['dates']); 
       $Q=$this->db->get('leave_reports'); 
       if($Q->num_rows() > 0) 
       { 
        foreach ($Q->result_array() as $row) 
        { 
         $data1[] = $row; 
        } 
       } 
       $column = 'leave_' . $date['dates']; 
       $employee[$column] = $data1[0]['leave_time'];//stored leave of each employee in this array 
       $data1=""; 
      } 
      $blk7[] = $employee; 
     }