如何使用php动态地将id传递给弹出窗口

Question

如何使用php动态地将id传递给弹出窗口？我需要自动将id值增加到弹出窗口.Plz帮我解决这个问题。     

include("config.php");

$query=mysql_query("select * from reports_form");
    while($row=mysql_fetch_array($query))
 {
 $id=$row['id'];
  $job_detail1 = $row['job_description']; ?>   
 <script>
 function myFunction() {
  var popup = document.getElementById("myPopup");
  popup.classList.toggle("show");
  }</script> 

       <td align="center"> <div class="popup" onclick="myFunction()">
       <a href="#myPopup<?php echo $id; ?>"> Click me!</a>

  <span class="popuptext"  id="myPopup"><?php echo "$job_detail1";?>            </span>
  </div></td>

    </tr>

  <?php
 }
 ?>

Answer 1

首先，你不应该在循环中使用你的脚本，并且传递你的函数变量它应该在

之下

?>
<script>
 function myFunction(id) {
  var popup = document.getElementById("myPopup");
console.log(id)
  popup.classList.toggle("show");
  }
</script> 
<?php
include("config.php");   
$query=mysql_query("select * from reports_form");
    while($row=mysql_fetch_array($query))
 {
 $id=$row['id'];
  $job_detail1 = $row['job_description']; ?> 

       <td align="center"> <div class="popup" onclick="myFunction(<?php echo $id;?>)">
       <a href="#myPopup<?php echo $id; ?>"> Click me!</a>

  <span class="popuptext"  id="myPopup"><?php echo "$job_detail1";?>    </span>
  </div></td>

</tr>

  <?php
 }
 ?>

第二个工作代码

<?php
//your prev code here
?>
<script>
    function myFunction(id) {
        var popup = document.getElementById("myPopup");
        console.log(id)
        popup.classList.toggle("show");
    }
</script>
    <table>
<?php

for($i=0;$i<10;$i++)
{
$id=$i;
$job_detail1 = 'desc'.$i; ?>

<td align="center"> <div class="popup" onclick="myFunction(<?php echo $id;?>)">
        <a href="#myPopup<?php echo $id; ?>"> Click me!</a>

        <span class="popuptext"  id="myPopup"><?php echo "$job_detail1";?>    </span>
    </div></td>

</tr>

<?php
}

?>
    </table>

截图