2017-07-28 80 views
0

以下是我正在處理的功能。我需要獲取while循環中的選定選項。 每次我試圖通過select id獲取選定的值時,它只會獲取行的其餘部分的最後一個選定值。獲取在嵌套循環中動態生成的選定選項值

請幫幫我。我如何獲取和存儲全表數據。

function generate_schedule($con,$sy,$year_id){ 

    $time = array(1=>'9:30-11:00',2=>'11:30-1:00',3=>'1:30-3:00',4=>'3:15-4:45'); 
    $days = array(1=>'Sunday',2=>'Monday',3=>'Tuesday',4=>'Wednesday',5=>'Thursday',6=>'Friday',7=>'Saturday'); 
    $rooms=array(1=>'302',2=>'303',3=>'304',4=>'701',5=>'702',6=>'704',7=>'901',8=>'902',9=>'1402',10=>'1405'); 
    // $yls = mysqli_query($con,"SELECT * from year_level_subject where year_id='$year_id'"); 
    $yls=mysqli_query($con,"SELECT * FROM year_level_subject as yls join subjects as s on yls.subj_id=s.subj_id where yls.year_id='$year_id'"); 
    $num=mysqli_num_rows($yls); 
    $x = 0; 
    echo '<table class="table table-bordered table-hover" id=" datatable">'; 
    echo '<thead> 
      <tr> 
      <th>Time</th> 
      <th>Days</th> 
      <th>Subject</th> 
      <th>Teacher</th> 
      <th>Room</th> 
      </tr> 
      </thead>'; 
    while($row=mysqli_fetch_assoc($yls)){ 
      extract($row); 
      echo '<tr>'; 
      echo '<td><select id ="seltime" ><option value=0>Select Time</option>'; 
      foreach($time as $key => $value) { 
      echo '<option value='.$key.'>'; 
      echo $value .'</option>'; 
      } 
      echo '</select></td>'; 
      echo '<td><select class="selday" ><option value=0>Select Day</option>'; 
      foreach($days as $key => $value) { 
      echo '<option value='.$key.'>'; 
      echo $value .'</option>'; 
      } 
      echo '</select></td>'; 
      echo '<td>'.$subj_desc.'</td>'; 
      $teach = mysqli_query($con,"SELECT * FROM teachers"); 
     echo '<td><select id="selteacher" onchange="schedata(this)" > 
       <option value=0>Select Teacher</option>'; 
     while($row=mysqli_fetch_assoc($teach)){ 
      extract($row); 
      echo '<option value='.$teach_id.'>'.$teach_fname.' '.$teach_mname.' '.$teach_lname.'</option>'; 
     } 
     echo '</select></td>'; 
      echo '</select></td>'; 
      echo '<td><select name ="selroom" ><option value=0>Select Room</option>'; 
      foreach($rooms as $key => $value) { 
      echo '<option value='.$key.'>'; 
      echo $value .'</option>'; 
      } 
      echo '</select></td>'; 

     echo '</tr>'; 
    } 
    echo '</table>';} 
+0

jQuery中或插入到數據庫 – Bhargav

回答

0

你我以前不給名稱屬性附加傷害到你選擇的元素

echo '<td><select id="selteacher" onchange="schedata(this)" > 

起名爲這是

echo '<td><select id="selteacher" name="selteacher" onchange="schedata(this)" >