我正在使用phonegap开发一个应用程序。我试图将mysql数据库连接到phonegap，以使用Ajax, php和jquery从html5页面保存数据。但是我无法在mySql数据库中保存数据。下面的代码有什么问题吗??下面是我的Html和php代码:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8" />
<meta name="format-detection" content="telephone=no" />
<meta name="viewport" content="user-scalable=no, initial-scale=1, maximum-scale=1, minimum-scale=1, width=device-width, height=device-height, target-densitydpi=device-dpi" />
<link rel="stylesheet" type="text/css" href="css/index.css" />
<title>MySQL Database with Phonegap</title>
<script type="text/javascript" src="cordova.js"></script>
<script type="text/javascript" src="js/index.js"></script>
<script src="//code.jquery.com/jquery-1.10.2.js"></script>
<script>
$('form').submit(function()
{
var postData = $(this).serialize();
$.ajax(
{
type: 'POST',
data: postData,
url: 'http://localhost/Save.php',
success: function(data)
{
console.log(data);
alert('Your data was successfully added');
},
error: function()
{
console.log(data);
alert('There was an error adding your comment');
}
});
return false;
});
</script>
</head>
<body>
<h1 align="center">Database Connectivity</h1><br><br><br>
<div align="center">
<form>
<label><b>First Name :</b></label><br>
<input type="text" id="fname" name="fname"/><br>
<label><b>Last Name :</b></label><br>
<input type="text" id="lname" name="lname"/><br>
<label><b>Email ID :</b></label><br>
<input type="email" id="email" name="email"/><br><br>
<input type="submit" value="SUBMIT"/>
<input type="reset" value="RESET"/>
</form>
</div>
</body>
</html>
Save.php
<?php
$host='localhost';
$uname='root';
$pwd='xyz';
$db='Sample';
$con=mysql_connect($host,$uname,$pwd) or die("Connection Failed");
mysql_select_db($db,$con) or die("database selection failed");
$fname = mysql_real_escape_string($_POST['fname']);
$lname = mysql_real_escape_string($_POST['lname']);
$email = mysql_real_escape_string($_POST['email']);
//echo "Hello";
$flag['code']=0;
if($r=mysql_query("INSERT INTO Login (FirstName, LastName, Email) VALUES ('".$fname."','".$lname."','".$email."')",$con))
{
$flag['code']=1;
echo "okk...";
}
print(json_encode($flag));
mysql_close($con);
?>
$.ajax({
url: "https://yourphpfilehere.php?PARAMETERSHERE=" + javascriptvariable,
success: function(res) {
alert(res);
}
});
，在你的php文件中，我建议你这样使用:
$myvariable = $_REQUEST['URLPARAMETER'];
//do the mysql statement,proccess here . . .
你应该从php脚本中使用$_REQUEST方法访问变量，因为你要在那个php文件上请求。不要让你的代码太复杂