我的HTML代碼是如何讓servlet的
<form action="/FirstServlet/Profile" method="post" class="login">
UserName: <input type="text" id="uName" name="name"><p>
Password: <input type="password" id="password" name="password"><br>
<input type="submit" id="button" value="login">
</form>
和JavaScript代碼是
<script src="js/jquery-3.2.1.min.js"></script>
<script src="js/jquery.serializeObject.js"></script>
<script src="js/Gruntfile.js"></script>
<script>
$(document).ready(function(){
var form=$('form.login').serializeObject();
alert(form);
form.submit(function(){
$.ajax({
type:form.attr('method'),
url:form.attr('action'),
dataType:'json',
data:Json.stringify(form),
success:function(data){
}
});
});
});
</script>
但我仍然得到答案這種格式
name=zeeshan&[email protected]
我想以json格式得到答案,如:
{"name":"zeeshan","password":"[email protected]"}
我越來越
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(request.getInputStream()));
String json = "";
if(br != null){
json = br.readLine();
}
System.out.println(json);
}
我搜索了很多數據的servlet代碼,每一個地方,他們說,這段代碼的輸出是JSON,但實際上我並沒有在該格式得到它。
如何讓我的輸出成爲JSON格式?
感謝哥們,這幫了我......我喜歡 –
data:{'json':JSON.stringify($(form).serializeObject())}, –