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Javascript 如何使用php将id动态传递给弹出窗口_Javascript_Php_Popupwindow - Fatal编程技术网

Javascript 如何使用php将id动态传递给弹出窗口

Javascript 如何使用php将id动态传递给弹出窗口,javascript,php,popupwindow,Javascript,Php,Popupwindow,如何使用php将id动态传递给弹出窗口?我需要自动增加id值弹出窗口。请帮助我解决这个问题。 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>

如何使用php将id动态传递给弹出窗口?我需要自动增加id值弹出窗口。请帮助我解决这个问题。
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
 }
 ?>
include(“config.php”);
$query=mysql\u query(“从报表中选择*);
while($row=mysql\u fetch\u array($query))
{
$id=$row['id'];
$job_detail1=$row['job_description'];?>
函数myFunction(){
var popup=document.getElementById(“myPopup”);
popup.classList.toggle(“显示”);
} 

首先,您不应该在循环中使用脚本,而应该将变量传递给下面的函数

?>
<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
 }
 ?>
?>
函数myFunction(id){
var popup=document.getElementById(“myPopup”);
控制台日志(id)
popup.classList.toggle(“显示”);
}

你确定你从PHP得到的是有效的Id吗?我需要根据相应的Id得到值。我只得到一个值。好的,我想我这边有一个误解,首先是语法错误js或PHP在哪里?你使用“Id”的目的是什么在这个代码中,所以这个code@FerhatBAŞ,感谢您的支持,.第一个比第二个好。我希望输出是根据onclick从db获得的id,然后列出该id的详细信息。第二个coe只是工作代码的证明,因为我没有让您的db只是为循环而更改,而不是您的while
<?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>