SpringMVC 通过ajax 前后端数据交互
在前端的开发过程中,经常在html页面通过ajax进行前后端数据的交互,SpringMVC的controller进行数据的接收,但是有的时候后端会出现数据无法接收到的情况,这个是因为我们的参数和前端ajax的contentType参数 类型不对应的情景,或者说contentType和后台controller 方法参数到底存在什么样的关系
普通的参数我们传递的时候往往是这样的两种情况:
contentType: "application/x-www-form-urlencoded",
contentType: "application/json",
这样的两种方式有什么样的区别,
第一种方式:application/x-www-form-urlencoded 参数会解析为参数表,比如:
-
name=John+Doe&age=30&city=New+York
通过ajax 传递,ajax写法如下:
let params={"username":"张三","password":"123456",}$.ajax({url: "dologin4",contentType: "application/x-www-form-urlencoded",headers: { 'Authorization': "****",'Access-Control-Allow-Origin':'*'},type: 'post',data: params,success: function(result) {console.log(result)},error: function(data) {console.log('接口不通');}});
这样的形式,后台接受的时候,使用:request.getParameter()
或@RequestParam,比如:
@RequestMapping("/dologin")public ModelAndView dologin(@RequestParam String username,@RequestParam String password) throws Exception {System.out.println(username);System.out.println(password);ModelAndView mav = new ModelAndView();mav.addObject("info", "欢迎你");mav.setViewName("success");return mav;}@ResponseBody@RequestMapping("/dologin2")public Map<String,Object> dologin2(HttpServletRequest request,HttpServletResponse response) throws Exception {String username=request.getParameter("username");System.out.println(username);Map<String,Object> map=new HashMap<>();map.put("aa", "1111");return map;}
适合 x-www-form-urlencoded
的情况:
-
传统HTML表单提交
-
简单的键值对数据
-
需要向后兼容老系统
-
文件上传(结合
multipart/form-data
)
第二种方式:contentType: "application/json", 整个body作为单一数据流处理,比如:
{"name": "John Doe","age": 30,"city": "New York","hobbies": ["reading", "swimming"] }
通过ajax 传递,ajax写法如下:
function dologin(){let params={"username":"张三","password":"123456",}$.ajax({url: "dologin4",contentType: "application/json",headers: { 'Authorization': "****",'Access-Control-Allow-Origin':'*'},type: 'post',data: JSON.stringify(params),success: function(result) {console.log(result)},error: function(data) {console.log('接口不通');}});}
Java后台接受前台传入参数的代码如下:@RequestBody
@ResponseBody@RequestMapping("/dologin4")public Map<String,Object> dologin4(HttpServletRequest request,HttpServletResponse response) throws Exception {String uu=request.getParameter("username");System.out.println(uu);StringBuilder jsonBuilder = new StringBuilder();try (BufferedReader reader = request.getReader()) {String line;while ((line = reader.readLine()) != null) {jsonBuilder.append(line);}}String jsonString = jsonBuilder.toString();ObjectMapper mapper = new ObjectMapper();Map<String, Object> jsonMap = mapper.readValue(jsonString, Map.class); String username = (String) jsonMap.get("username");System.out.println(username);Map<String,Object> map=new HashMap<>();map.put("aa", "1111");return map; }@ResponseBody@RequestMapping("/dologin3")public Map<String,Object> dologin3(@RequestBody Map<String,Object> reqMap) throws Exception {String username=reqMap.get("username").toString();System.out.println(username);Map<String,Object> map=new HashMap<>();map.put("aa", "1111");return map;}
适合 application/json
的情况:
-
RESTful API通信
-
复杂数据结构
-
需要明确数据类型
-
前后端分离架构
-
移动应用与服务器通信
两种情景是不一样的,如果你前端传入的是json格式,那么后端你用:
String uu=request.getParameter("username");
System.out.println(uu);
这样是无法接收到数据的,接收到的参数一直为null,因为json是整体作为一个流传入到后台的
希望对你有所帮助!