您可以将PhoneGap 与PHP 和jQuery Ajax 一起使用来获取内容。在文件头加载 jQuery 库。在函数onBodyLoad(), 处为 PHP 文件调用 Ajax:
$('#content').load('http://www.example.com/test.php');
在 HTML 会话中,将 div id="content" 放在要显示内容的位置。
PHP:
for($i=1; $i<=10; $i++) {
echo '<p>Dinamic content coming from test.php! Value: ' . $i . ' of 10.</p>';
}
HTML 将打印:
<p>Dinamic content coming from test.php! Value: 01 of 10.</p>
<p>Dinamic content coming from test.php! Value: 02 of 10.</p>
<p>Dinamic content coming from test.php! Value: 03 of 10.</p>
<p>Dinamic content coming from test.php! Value: 04 of 10.</p>
<p>Dinamic content coming from test.php! Value: 05 of 10.</p>
<p>Dinamic content coming from test.php! Value: 06 of 10.</p>
<p>Dinamic content coming from test.php! Value: 07 of 10.</p>
<p>Dinamic content coming from test.php! Value: 08 of 10.</p>
<p>Dinamic content coming from test.php! Value: 09 of 10.</p>
<p>Dinamic content coming from test.php! Value: 10 of 10.</p>
要将内容发送到另一个页面并让用户登录,您可以这样做
$.get('login.php?name=user', function(data) {
$('#content').html(data);
});
你的 login.php 可能有类似的东西:
if (isset($_GET['name'])) {
$name = $_GET['name'];
echo "Name: $name";
} else {
echo "Please enter a valid name!!";
}
为了确保您的登录安全,您可以使用 POST 方法,如下所述:
$('#form').submit(function() {
$.post('login.php', $('#form').serialize(), function(data) {
$('#content').html(data);
});
return false; // to avoid page going to login.php file
});
还有login.php
if(!empty($_POST)) {
$user = $_POST['name'];
$pass = $_POST['password'];
// db query and stuff goes here...
echo "Worked!";
} else {
"Enter values!";
}