Question

我想从远程数据库中检索信息并将其显示在列表中。什么是最好的方法？目前我正在调用一个PHP但无法使其工作。有什么建议吗？提前致谢

---- ----的index.html

<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="width=device-width, minimum-scale=1, maximum-scale=1">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8"/> 
<title>Index</title>
    <link href="css/custom.css" rel="stylesheet" type="text/css" />
   <link rel="stylesheet" href="http://code.jquery.com/mobile/1.0/jquery.mobile-    1.0.min.css" />
   <script type="text/javascript" src="http://code.jquery.com/jquery-1.6.4.min.js"></script>
   <script type="text/javascript" src="http://code.jquery.com/mobile/1.0/jquery.mobile-1.0.min.js"></script>
    <script src="cordova-2.6.0.js"></script>
    <script>
    $(document).ready(function() {


            $("#resultBlock").html("Waiting for data...");

            $.get("http://localhost/select.php", {}, function(data) {

                $("#resultBlock").html(data);
            })
    });
    </script>
</head>
<body>
<div data-role="page" id="inicio">
      <div data-role="header" data-position="fixed">
        <h1>Periódico</h1>
      </div>
       <div data-role="content">

            <p id="resultBlock"></p>
       </div>
       <div data-role="footer" data-id="foo1" data-position="fixed" >
        <div data-role="navbar">
            <ul>
                <li><a href="" class="ui-btn-active ui-state-persist"><img src="images/periodico-icon.png" width="40" height="40"><br/>Periodico</a></li>
                <li><a href="#directorio" ><img src="images/directorio-icon.png" width="40" height="40"><br/>Directorio</a></li>
                <li><a href="#clasificados" ><img src="images/clasificados-icon.png" width="40" height="40"><br/>Clasificados</a></li>
                 <li><a href="#cuponera" ><img src="images/cuponera-icon.png" width="40" height="40"><br/>Cuponera</a></li>
            </ul>
        </div>
      </div>
</div>
</body>
</html>

- - - - - - - - select.php

<?php

$con=mysqli_connect("localhost","localhost","","test");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$result = mysqli_query($con,"SELECT * FROM Persons");

echo "<table border='0'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
</tr>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['FirstName'] . "</td>";
  echo "<td>" . $row['LastName'] . "</td>";
  echo "<td>" . $row['Age'] . "</td>";
  echo "</tr>";
  }

echo "</table>";
mysqli_close($con);
?>

Answer 1

我记得在使用phonegap进行开发时遇到了jquery ajax请求的问题。尝试使用经典的ajax请求，它对我有用。另一个可能的问题是您正在请求计算机上的本地文件。示例代码如下：

function loadContent(){

    var xmlhttp = getXmlHttpObject();               

    xmlhttp.onreadystatechange=function()
      {          
      if (xmlhttp.readyState==4 && xmlhttp.status==200)
        {
            document.getElementById("listContent").innerHTML=xmlhttp.responseText;

        }
      };
    var url = "http://example.com/select.php";
    xmlhttp.open("GET",url,true);
    xmlhttp.send();

获取xmlhttp对象：

function getXmlHttpObject(){
    var xmlhttp;

if (window.XMLHttpRequest)        
    xmlhttp=new XMLHttpRequest();         
else        
    xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");     

return xmlhttp;

}

从PhoneGap调用PHP

1 个答案: