所以我有这个HTML表单:
<html>
<head><title>test</title></head>
<body>
<form action="myurl" method="POST" name="myForm">
<p><label for="first_name">First Name:</label>
<input type="text" name="first_name" id="fname"></p>
<p><label for="last_name">Last Name:</label>
<input type="text" name="last_name" id="lname"></p>
<input value="Submit" type="submit" onclick="submitform()">
</form>
</body>
</html>
当用户点击提交时,这是将此表单的数据作为JSON对象发送到我的服务器的最简单方法？
更新:我已经走了这么远,但它似乎不起作用:
<script type="text/javascript">
function submitform(){
alert("Sending Json");
var xhr = new XMLHttpRequest();
xhr.open(form.method, form.action, true);
xhr.setRequestHeader('Content-Type', 'application/json; charset=UTF-8');
var j = {
"first_name":"binchen",
"last_name":"heris",
};
xhr.send(JSON.stringify(j));
我究竟做错了什么？
我正在使用spring mvc.当method = post时,我无法从url获取param.但是当我将方法改为GET时,我可以得到所有的参数.
这是我的表格:
<form method="POST" action="http://localhost:8080/cms/customer/create_customer" id="frmRegister" name ="frmRegister" enctype="multipart/form-data">
<input class ="iptRegister" type="text" id="txtEmail" name="txtEmail" value="" />
<input class ="iptRegister" type="password" id="txtPassword" name="txtPassword" value="" />
<input class ="iptRegister" type="text" id="txtPhone" name="txtPhone" value="" />
<input type="button" id="btnRegister" name="btnRegister" value="Register" onclick="" style="cursor:pointer"/>
</form>
这是我的控制器:
@RequestMapping(value= "/create_customer", method = RequestMethod.POST)
@ResponseBody
public String createCustomer(HttpServletRequest request,
@RequestParam(value="txtEmail", required=false) String email,
@RequestParam(value="txtPassword", required=false) String password,
@RequestParam(value="txtPhone", required=false) String phone){
ResultDTO<String> rs = new ResultDTO<String>();
rs.setStatus(IConfig.SHOW_RESULT_SUCCESS_ON_MAIN_SCREEN);
try{
Customer c = new Customer();
c.setEmail(email); …之间有什么根本区别@RequestParam和@QueryParam