1.指定为 post 请求，输入 url
Content-Type: application/x-www-form-urlencoded;charset=utf-8
request body中的参数格式：userName=adminicxp&userPassword=123qwe!@#
这种方式可以用 request.getParameter的方式来获得。
2.指定为 post 请求，输入 url
Content-Type: application/json; charset=utf-8
request body中的参数格式：
{"userName": "adminicxp","userPassword": "123qwe!@#","sysId": "xxx" }
这种方式通过如下方式获得：
@RequestMapping("/xxx") @ResponseBody public String xxx(HttpServletRequest request) throws IOException { String jsonString = getBodyString(request.getReader()); JSONObject jbJsonObject = new JSONObject().fromObject(jsonString); User user = (User) JSONObject.toBean(jbJsonObject, User.class); System.out.println(jbJsonObject); System.out.println("id:" + user.getUserName());return null; } @RequestMapping("/xxx2") @ResponseBody public String xxx2(User user) throws IOException { System.out.println("---------------"); System.out.println(user.getUserName()); System.out.println(user.getPassWord()); System.out.println("---------------"); if (true) { return "success"; } else { return "fail"; } }
public String getBodyString(BufferedReader br) {
String inputLine;
String str = "";
try {
while ((inputLine = br.readLine()) != null) {
str += inputLine;
}
br.close();
} catch (IOException e) {
System.out.println("IOException: " + e);
}
return str;
}
3.post数组
方式1：
springmvc 后台java代码
@RequestBody Map<String, List<String>> paramList<String> ids = param.get("ids"); fiddler 请求
Content-Type指定为 application/jsonRequestBody格式：{"ids":["first0001company", "xxx4234324"]}
方式2：
springmvc 后台java代码
@RequestParam List<String> ids;或者 @RequestParam String[] ids;
fiddler 请求
Content-Type指定为 application/x-www-form-urlencodedRequestBody格式：ids=first0001company&ids=xxx4234324
