Question

我正在使用SpringMVC，我将数据从ajax传递给控制器，但我的控制器中的值为null，请检查下面的代码

function searchText()
{
   var sendData = {
    "pName" : "bhanu",
     "lName" :"prasad"
   }
  $.ajax({
type: "POST",
 url: "/gDirecotry/ajax/searchUserProfiles.htm,
    async: true,
    data:sendData,
    success :function(result)
    {
    }
 }

MyControllerCode

         RequestMapping(value="/gDirecotry/ajax/searchUserProfiles.htm",method=RequestMethod.POST)

       public  @ResponseBody String  getSearchUserProfiles(HttpServletRequest request)
       {
         String pName = request.getParameter("pName");
         //here I got null value
       }

任何人帮助我

Answer 1

嘿，请享受以下代码。

Javascript AJAX调用

function searchText() {
   var search = {
      "pName" : "bhanu",
      "lName" :"prasad"
   }
   $.ajax({
      type: "POST",
      contentType : 'application/json; charset=utf-8',
      dataType : 'json',
      url: "/gDirecotry/ajax/searchUserProfiles.htm",
      data: JSON.stringify(search), // Note it is important
      success :function(result) {
       // do what ever you want with data
     }
  });

}

Spring控制器代码

RequestMapping(value="/gDirecotry/ajax/searchUserProfiles.htm",method=RequestMethod.POST)

   public  @ResponseBody String  getSearchUserProfiles(@RequestBody Search search, HttpServletRequest request) {
       String pName = search.getPName();
       String lName = search.getLName();

       // your logic next
   }

以下搜索课程如下

class Search {
    private String pName;
    private String lName;

    // getter and setters for above variables
}

此类的优点在于，将来您可以根据需要向此类添加更多变量。
例如，如果您想实现排序功能。

Answer 2

如果您不想创建一个类，可以直接发送JSON而不使用Stringifying，请使用此方法。使用默认内容类型。只需使用@RequestParam代替@RequestBody即可。 @RequestParam名称必须与json中的名称相同。

Ajax Call

function searchText() {
    var search = {
    "pName" : "bhanu",
    "lName" :"prasad"
    }
    $.ajax({
    type: "POST",
    /*contentType : 'application/json; charset=utf-8',*/ //use Default contentType
    dataType : 'json',
    url: "/gDirecotry/ajax/searchUserProfiles.htm",
    data: search, // Note it is important without stringifying
    success :function(result) {
    // do what ever you want with data
    }
    });

Spring Controller

RequestMapping(value="/gDirecotry/ajax/searchUserProfiles.htm",method=RequestMethod.POST)

   public  @ResponseBody String  getSearchUserProfiles(@RequestParam String pName, @RequestParam String lName) {
       String pNameParameter = pName;
       String lNameParameter = lName;

       // your logic next
   }

Answer 3

我希望，你需要包含dataType选项，

dataType: "JSON"

您可以在控制器中获取表单数据，如下所示

 public  @ResponseBody String  getSearchUserProfiles(@RequestParam("pName") String pName ,HttpServletRequest request)
       {
         //here I got null value
       }

Answer 4

如果您可以设法在一个查询字符串参数中传递整个json，那么您可以在服务器端使用rest模板从json生成Object，但仍然不是最佳方法

Answer 5

u take like this 
var name=$("name").val();
var email=$("email").val();

var obj = 'name='+name+'&email'+email;
  $.ajax({
   url:"simple.form",
   type:"GET",
   data:obj,
   contentType:"application/json",
   success:function(response){
  alert(response);
  },
  error:function(error){
  alert(error);
  }
});

spring Controller

@RequestMapping(value = "simple", method = RequestMethod.GET)
public @ResponseBody String emailcheck(@RequestParam("name") String name, @RequestParam("email") String email, HttpSession session) {

    String meaaseg = "success";
    return meaaseg;
}

如何将Json对象从ajax传递给spring mvc控制器？

5 个答案: