为动态html表中的一列创建下载链接

时间:2018-07-19 09:28:56

标签: php mysql ajax

我有一个来自数据库的表。我想为

创建下载
  

恢复列

在表中,它应该下载或查看该特定的行文件。我该怎么办。

enter image description here

jobs.php

     <div class="col-md-8 col-md-offset-2">
    <h3 style="text-align: center">Job Application Submitted</h3>
    <table>
  <thead>
  <tr class="table-headers">
    <th>Name</th>
    <th>Email</th>
    <th>Phone</th>
    <th>job Applied For</th>
    <th>Resume</th>
  </tr>
  </thead>
      <tbody>
<?php include('config.php');


  $sql = "SELECT * FROM apply_job order by id ASC";
  $result = mysqli_query($db,$sql);
  if(mysqli_num_rows($result) > 0)
  {
   while($row = mysqli_fetch_assoc($result))

  {

?>    


    <tr>
      <td><?php echo $row['template-jobform-fname']?></td>
      <td><?php echo $row['template-jobform-email'];?></td>
      <td><?php echo $row['template-jobform-phone'];?></td>
       <td><?php echo $row['job_name'];?></td>
     <td><a href="<?php echo $row['template-jobform-application']?>" target="_blank" ><?php echo $row['template-jobform-application'];?></a></td>

    </tr>
   <?php 
}
}
?>
  </tbody>
</table>
</div>

我想要下载特定文件,即如果我更新第二行并下载第二行简历,则应该下载该特定简历而不是全部。我该怎么办。

2 个答案:

答案 0 :(得分:1)

Use sample function below for file download
<?php
function fileDownload($filename, $filepath)
{
    $filepath = $filepath . "/" . $filename;
    header("Cache-control: private");
    header("Content-type: application/force-download");
    header("Content-transfer-encoding: binary\n");
    header("Content-disposition: attachment; filename=\"$filename\"");
    header("Content-Length: " . filesize($filepath));
    readfile($filepath);
}

答案 1 :(得分:0)

您可以这样创建下载链接:

<a href="<?php echo $row['template-jobform-application']?>" download ><?php echo $row['template-jobform-application'];?></a>

请注意,标记定义中存在download属性,该属性指定应在单击时下载目标。