我创建了一个视图和控制器,我试图使用jquery将表单参数传递给控制器..
$.ajax({
type: "POST",
url: "add.htm",
datatype: "json",
data: "name=" + name + "&password=" + password + "&gender=" + gender + "&aboutyou=" + aboutyou,
success: function(response){
alert(response.name);
// we have the response
if(response.status == "SUCCESS"){
$('#info').html("User has been added to the list successfully");
/* $('#name').val('');
$('#education').val(''); */
}else{
$('#info').html("Sorry, there is some thing wrong with the data provided.");
}
},
error: function(e){
alert('Error: ' + e);
}
});
}
这是我的控制器。
public ModelAndView add(HttpServletRequest request,
HttpServletResponse response, employee employee) throws Exception {
List list=new ArrayList();
employeedao.saveUser(employee);
return new ModelAndView("userform");
}
答案 0 :(得分:1)
我认为您想要获取控制器中的值(来自问题标题),如果这是要求,则使用request.getParameter("parametername");
例如,要获取名称,请执行此操作
request.getParameter("name");
您还需要更改传递参数的格式,如
data:{parametername:parametervalue}
作为JSON
示例data:{name:name ....}
答案 1 :(得分:0)
替换您的以下行: -
data: "name=" + name + "&password=" + password + "&gender=" + gender + "&aboutyou=" + aboutyou,
以下行: -
data: { name:name,password:password,gender:gender,aboutyou:aboutyou}
答案 2 :(得分:0)
答案 3 :(得分:0)
您已经像普通数据一样发送了数据,但在jquery中,格式的变化如下所述
$.ajax({
type: "POST",
url: "add.htm",
datatype: "json",
data: { "name":name,"password":password,"gender":gender,"aboutyou":aboutyou},
success: function(response){
alert(response.name);
// we have the response
if(response.status == "SUCCESS"){
$('#info').html("User has been added to the list successfully");
/* $('#name').val('');
$('#education').val(''); */
}else{
$('#info').html("Sorry, there is some thing wrong with the data provided.");
}
},
error: function(e){
alert('Error: ' + e);
}
});
}
这是发送数据的格式。 谢谢 纳温