Ajax无法在Phonegap项目中工作，但可以在外部运行吗？

Question

此刻，我正在创建一个PhoneGap应用程序，并且我一直在使用ajax在该应用程序的html页面中显示来自我的MySQL数据库的信息。如果我在PhoneGap外部运行HTML页面，它将显示该表，但它不在PhoneGap中，而且我不确定为什么我对PhoneGap和编码工作还是很陌生？

我的代码如下；

display.php：

<?php
include("connection.php");
mysqli_select_db($con, "mscim2018_lmm12");
$result=mysqli_query($con, "select * from food");

echo "<table border='1' >
<tr>
<td align=center> <b>Roll No</b></td>
<td align=center><b>Name</b></td>
<td align=center><b>Address</b></td>
<td align=center><b>Stream</b></td></td>
<td align=center><b>Status</b></td>";

while($data = mysqli_fetch_row($result))
{   
    echo "<tr>";
    echo "<td align=center>$data[0]</td>";
    echo "<td align=center>$data[1]</td>";
    echo "<td align=center>$data[2]</td>";
    echo "<td align=center>$data[3]</td>";
    echo "<td align=center>$data[4]</td>";
    echo "</tr>";
}
echo "</table>";
?>

connection.php：

<?php 
$con=mysqli_connect("--","--","--"); 
if (!$con)
{
    die('Could not connect: ' . mysqli_error());
}   
?>

（正在使用framework7，因此没有HTML标记，但是脚本是在索引文件中准确地提供的）food.html;

<div class="navbar">
   <div class="navbar-inner">
      <div class="left">
         <a href="index.html" class="back link">
         <i class="icon icon-back"></i>
         <span>Back</span>
         </a>
      </div>
      <div align="center" class="center sliding"><strong>CFPal - BMR</strong></div>
      <div class="right">
         <a href="#" class="link icon-only open-panel"><i class="icon icon-bars"></i></a>
      </div>
   </div>
</div>
<div class="pages" align="center">
   <div data-page="about" class="page" align="center">
      <div class="page-content">
         <div class="content-block" id="bmipage">
  
  
  
  <h3 align="center">Manage Student Details</h3>
<table border="1" align="center">
   <tr>
       <td> <input type="button" id="display" value="Display All Data" /> </td>
   </tr>
</table>
<div id="responsecontainer" align="center">
  
  </div>
  
         </div>
      </div>
   </div>
</div>

和ajax.js：

 $(document).ready(function() {

    $("#display").click(function() {                

      $.ajax({    //create an ajax request to display.php
        type: "GET",
        url: "https://cs1.ucc.ie/~lmm12/ProjectTraining/display.php",             
        dataType: "html",   //expect html to be returned                
        success: function(response){                    
            $("#responsecontainer").html(response); 
            //alert(response);
        }

    });
});
});