See*_*u69 4 javascript java ajax jquery
refer.jvmhost.net/refer247/registration,这是我的网址,我必须获取这个网址的请求,如用户详细信息,json如果它包含..Dont给我android代码,应该以格式状态n错误获取相应的响应.
这是html页面.
<head>
<script type="text/javascript" src="json2.js"></script>
</head>
<body>
<div data-role="page" data-theme="c">
<div data-role="header" data-position="fixed" data-inset="true" class="paddingRitLft" data-theme="c">
<div data-role="content" data-inset="true"> <a href="index.html" data-direction="reverse"><img src="images/logo_hdpi.png"/></a>
</div>
</div>
<div data-role="content" data-theme="c">
<form name="form" method="post" onsubmit="return validate()">
<div class="logInner">
<div class="logM">Already have an account?</div>
<div class="grouped insert refb">
<div class="ref first">
<div class="input inputWrapper">
<input type="text" data-corners="false" class="inputrefer" placeholder="Userid" name="userid" id="userid" />
</div>
<div class="input inputWrapper">
<input type="password" data-corners="false" class="inputrefer" placeholder="Password" name="password" id="password" />
</div> <a href="dash.html" rel="external" style="text-decoration: none;"><input type="submit" data-inline="true" value="Submit" onclick="json2()"></a>
<p><a href="#" style="text-decoration: none;">Forgot Password</a>
</p>
</div>
</div>
<div class="logM">New user? Create refer Account</div>
<input type="button" class="btnsgreen" value="Sign Up! its FREE" class="inputrefer" data-corners="false" data-theme="c" />
</form>
</div>
</div>
<p style="text-align: center;">© refer247 2013</p>
</div>
</body>
这是json2.js
function json2()
{
var json1={"username":document.getElementById('userid').value,
"password":document.getElementById('password').value,
};
//var parsed = jsonString.evalJSON( true );
alert(json1["username"]);
alert(json1["password"]);
};
所以告诉我如何将json数据发送到该url n获取一些响应,如果你已经存在电子邮件id已存在,如果你注册了该id ..然后给出一些错误,如电子邮件ID已经存在n如果成功注册然后给予respone就像注册成功和状态msg..200 okk ...
您可以使用ajax将json数据发布到指定的url/controller方法.在下面的示例中,我发布了一个json对象.您也可以单独传递每个参数.
var objectData =
{
Username: document.getElementById('userid').value,
Password: document.getElementById('password').value
};
var objectDataString = JSON.stringify(objectData);
$.ajax({
type: "POST",
url: "your url with method that accpects the data",
dataType: "json",
data: {
o: objectDataString
},
success: function (data) {
alert('Success');
},
error: function () {
alert('Error');
}
});
并且您的方法只能有一个字符串类型的参数.
[HttpPost]
public JsonResult YourMethod(string o)
{
var saveObject = Newtonsoft.Json.JsonConvert.DeserializeObject<DestinationClass>(o);
}
小智 1
$.ajax({
url: urlToProcess,
type: httpMethod,
dataType: 'json',
data:json1,
success: function (data, status) {
var fn = window[successCallback];
fn(data, callbackArgs);
},
error: function (xhr, desc, err) {
alert("error");
},
});
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