尝试使用发送 Ajax 请求,像这样。我假设您将 php 用于动态代码(服务器端)。
这里是您的 cordova, phonegap 目录中的 HTML 文件示例。
<form method = "post" action = "#!">
<div class="col-md-4">
<span class="help-block">Name</span><input type="text" name="username" class="form-control" />
</div>
<br>
<div class="col-md-4">
<span class="help-block">Password</span><input type="text" name="password" class="form-control" />
</div>
<input type = "submit" value = "Save" class = "btn btn-success right" onClick="UpdateRecord();"/>
</form>
<script>
function UpdateRecord()
{
var name = $("[name='username']").val();
var host = $("[name='password']").val();
jQuery.ajax({
type: "POST",
url: "php/login.php",
/* Or */
/*url: "https://www.yoursite.com/page",*/
data: "username="+ username+"& password="+ password,
dataType: "html",
cache: false,
success: function(response){
if(response == 'true') {
$.session.set("myVar", username);
window.location.href='profile.html';
}
else {
$("#errorMessage").html("Invalid Entry, Please Try Again");
}
}
});
}
</script>
以及用于句柄查询的 PHP 文件。
请注意,代码未经测试,可能会根据您的需要进行更改。您可以在此处执行任何加密方法并使用任何功能。
<?php
include 'config.php';
$username = mysql_real_escape_string($_POST['username']);
$password = mysql_real_escape_string($_POST['password']);
if(!empty($username) && !empty($password))
{
//$result = mysql_query("SELECT * FROM ".$db.".users WHERE username='$username' and password ='$password'");
$result=mysql_query("select * from ".$db.".users WHERE email = '$username' ");
while($data = mysql_fetch_row($result))
{
$original_password = $data[3];
$salt = $data[4];
$hashedPass = sha1($salt.$password);
$fullusername = $data[16]." ".$data[17]; // Used Only for create full name session
if ($original_password == $hashedPass)
{
$_SESSION['username'] = $fullusername;
$_SESSION['useremail'] = $username;
$_SESSION['UserID'] = $data[0];
echo 'true';
}
}
}
?>
编辑
request.open("GET", 'http://url/service?firstElement='+elem+'&secondElement='+elem2, false);
在发送敏感数据时避免使用 GET 方法。
编辑,有用的链接
Local storage protection in phonegap application