关于JQuery用AJAX请求,从前台获取数据,出入后台,再从后台返回数据给前台
我不知道怎么写,我把我知道的放在这,谁知道的,给点指点,我现在在看选择器,同时也看看ajax的请求和事件,
有知道的,给我提几句 方向的话题也行,真的是迷茫啊。
<script type='text/javascript' src='js/jquery-1.4.4.js'></script>
<script type='text/javascript'>
var username;
var password;
$(document).ready(function() {
$.ajax( {
type : "POST",
url : "/login",
data : "name="+username+"&password="+password",
success : function(msg) {
alert("url hasPost!");
}
});
});
formSubmit(){
}
</script>
<body>
<form action="/login" method="post">
nausername:<input type="text" name="name" id="name" /><br/>
password:<input type="password" name="password" id="password"/><br/>
<input type="submit" value="提交"><input type="reset" value="重置"/>
</form>
</body>
<script type="text/javascript" src="js/jquery-1.4.4.js"></script>
<script type="text/javascript">
$(function(){
$(":input[type=button]").click(function(){
var $params="username="+$("#name").val()+"&password="+$("#password").val();
$.ajax({
type : "POST",
url: "TestAjaxServlet",
data: $params,
success: function(msg){
alert(msg);
}
});
});
})
</script>
</head>
<body>
<form action="" method="post">
nausername:<input type="text" name="name" id="name" /><br/>
password:<input type="password" name="password" id="password"/><br/>
<input type="button" id="submit" value="提交"><input type="reset" value="重置"/>
</form>
</body>
TestAjaxServlet
public void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
response.setContentType("text/html");
PrintWriter out = response.getWriter();
String username=request.getParameter("username");
String password=request.getParameter("password");
if(username.equals("hello")&&password.equals("world")){
out.print("success");
}else{
out.print("error");
}
out.flush();
out.close();
}
这下应该会了吧,不知道你是哪出了问题。
die(json_encode($arr));//后台向js传送的数据是$arr
?>
[解决办法]
本楼也在学呀...,看看load效果怎么样。
[解决办法]
$.get(url,param:params,CallBack)