如何在更新表時刷新標籤數據？

的index.php如何在更新表時刷新標籤數據？

<div class="tab"> 
    <button class="tablinks" onclick="openCity(event, 'Engineering')">Engineering</button> 
    <button class="tablinks" onclick="openCity(event, 'LAW')">LAW</button> 
</div> 

<div id="Engineering" class="tabcontent"> 
    <table class="items"> 
     <tr> 
      <th>State</th> 
      <th>College Name</th> 
     </tr> 
      <?php 
       $query = "select * from college where field = 'engineering'"; 
       $show = mysqli_query($link,$query); 
       while ($fetch = mysqli_fetch_array($show)) 
       { 
      ?> 
      <tr> 
       <td><?php echo $fetch['state']?></td> 
       <td><?php echo $fetch['college_name']?></td> 
       <td> 
        <a href="edit.php?id=<?php echo $fetch['id']; ?>">edit</a> 
       </td> 
      </tr> 
      <?php  
       } 
      ?> 
    </table> 
</div> 
<div id="Law" class="tabcontent"> 
    <table class="items"> 
     <tr> 
      <th>State</th> 
      <th>College Name</th> 
     </tr> 
      <?php 
       $query = "select * from college where field = 'law'"; 
       $show = mysqli_query($link,$query); 
       while ($fetch = mysqli_fetch_array($show)) 
       { 
      ?> 
      <tr> 
       <td><?php echo $fetch['state']?></td> 
       <td><?php echo $fetch['college_name']?></td> 
       <td> 
        <a href="edit.php?id=<?php echo $fetch['id']; ?>">edit</a> 
       </td> 
      </tr> 
      <?php  
       } 
      ?> 
    </table> 
</div>

edit.php

<?php 
if(isset($_POST['update'])) 
{ 
    $college_name = $_POST['colleges']; 
    $state = $_POST['state']; 
    $sqli = "update college set college_name = '$college_name', state = '$state' where id = '$id'"; 
    $results = mysqli_query($link,$sqli); 
    if($result == true) 
    { 
     $msg .= "<p style='color:green;'>Your data update successfully</p>"; 
    } 
    else 
    { 
     $msg .= "<p style='color:red;'>Errror!</p>"; 
    } 
} 
?> 
<form method="POST" enctype="multipart/form-data" > 
    <select name="state" id="state"> 
     <option value="<?php echo $stateid; ?>"><?php echo $statename; ?></option> 
     <option value="">Select State</option> 
     <?php 
     $sql = "select * from statemaster"; 
     $result = mysqli_query($link,$sql); 
     while($row = mysqli_fetch_array($result)) 
     { 
     echo "<option value=".$row['stateid'].">".$row['statename']."</option>"; 
     } 
     ?> 
    </select> 

    <select name="colleges" id="colleges"> 
     <option value="<?php echo $college_name; ?>"><?php echo $college_name; ?></option> 
     <option value="">Select College</option> 
    </select> 

    <button type="submit" name='update' id='update'>update</button> 
</form>

在這段代碼當我點擊編輯按鈕，然後就會去edit.php網頁，我得到的ID從URL和運行更新查詢更新表大學後，數據將更新，但是當我從編輯頁面移動到index.php頁面時，數據將保持不變，但數據庫更新數據將在那裏。那麼，我該如何解決這個問題？

謝謝

來源

2017-06-14 omkara

哪裏是你的編輯形式？您的更新查詢是否有效？ –

是的，@Fairy Dancer – omkara

來聊聊天就解釋清楚了。 http://chat.stackoverflow.com/rooms/146504/codeigniter –

檢查緩存。它可能是瀏覽器不會去服務器以獲取index.php的內容，因爲它認爲它有它。嘗試調用的index.php與變量，如：

<a href="index.php?<?=microtime()?>">Home</a>

來源

2017-06-14 07:20:06

或者你可以在你的瀏覽器中禁用緩存。 –

如何在更新表時刷新標籤數據？

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