Question

springmvc前台ajax中url怎么写

var json = {"username":username,"password":password};
alert("<%=request.getContextPath()%>/userLoginController/userLoginCheck");
$.ajax({
url:'userLoginController/userLoginCheck',
data:json,
type:'post',
cache : false,
dataType : 'json',
success:function(result){
if(result=="false"){
alert("用户名或密码输入有误");
return false;
}
if(result=="true"){
alert("通过验证");
return true;
}
},
error : function() {
alert("系统错误,请联系管理员!");
}
});
其中 url:'userLoginController/userLoginCheck', 里面写的是后台controller里面的方法名 但是死活进不去方法 还一直报错 "系统错误,请联系管理员!") 求解！！！ 没有财富值了。。。
@Controller
@RequestMapping("/userLoginController")
public class UserLoginController {
@Autowired
private UserLoginService userLoginService;

private static final String VIEW = "menu/mainMenu";

@RequestMapping(value="/userLoginCheck", method={RequestMethod.GET, RequestMethod.POST})
public String userCheck(HttpServletRequest request, HttpServletResponse response){
String username = request.getParameter("username");
String password = request.getParameter("password");
}

Answer 1

首先神行如。类上面
@Controller
@RequestMapping("/") --这里加访问路径
public class LoginController ---

方游启法--

@RequestMapping(value = "/usercheck", method = {RequestMethod.POST})
@ResponseBody
public String userLoginCheck (String param(说明：接受参数), HttpServletRequest req)
{
}

ajax的url可以直接写 url:'带毁usercheck',

追问

我的control是这样写的：
@Controller
@RequestMapping("/userLoginController")
public class UserLoginController {

@RequestMapping(value="/userLoginCheck", method={RequestMethod.GET, RequestMethod.POST})

追答

可以是你传参数错误
AJAX这样传参数
data:{
username:username,
password: password
},

@RequestMapping(value="/userLoginCheck", method={RequestMethod.GET, RequestMethod.POST})
public String userCheck(String username, String password, HttpServletRequest request, HttpServletResponse response){

}