嗨,我是Swift的新手,正在尝试制作一个简单的应用程序。
我正在使用“ alamofire 5 beta 6”发出请求。
下面是一些代码
-发出发布请求的代码
var json:JSON = JSON(["id":id.text, "password":enteredPassword])
var parameters: Parameters = ["id":id.text, "password":enteredPassword]
let headers:HTTPHeaders = [ "Content-Type":"application/json"]
AF.request("http://127.0.0.1:8080/user", method: .post, parameters: parameters, encoding: URLEncoding.httpBody, headers: headers).responseJSON{
response in
print("response : \(response)")
}
-Spring框架代码
@RequestMapping(value="user", method=RequestMethod.POST)
public JSONObject addUser(
@RequestBody Memberinfo member,
HttpServletRequest request) {
JSONObject result = new JSONObject();
return result;
}
-Memberinfo.java,在控制器中用于检索@RequestBody
public class Memberinfo {
String id;
String password;
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public String getPassword() {
return password;
}
public void setPassword(String password) {
this.password = password;
}
}
在Swift代码中,我设置了参数ID和密码以在Spring框架中将其取回。
但是,在我发出请求后,Alamofire便显示了消息
response : success({
error = "Bad Request";
message = "JSON parse error: Unrecognized token 'id': was expecting ('true', 'false' or 'null'); nested exception is com.fasterxml.jackson.core.JsonParseException: Unrecognized token 'id': was expecting ('true', 'false' or 'null')\n at [Source: (PushbackInputStream); line: 1, column: 4]";
path = "/user";
status = 400;
timestamp = "2019-06-09T05:46:07.417+0000";
})
答案 0 :(得分:0)
我认为应按以下方式发送参数
var parameters: Parameters = {"id":id.text, "password":enteredPassword}
另一个重要的事情,您不需要将JSONObject作为端点的响应类型,您只需使用@ReponseBody注释端点即可获得JSON响应
@RequestMapping(value="user", method=RequestMethod.POST)
@ResponseBody
public ResponseEntity<JSONObject> addUser(
@RequestBody Memberinfo member,
HttpServletRequest request) {
JSONObject result = new JSONObject();
return result;
}
在从外部源调用此终结点之前,始终使用Postman或RestClient之类的东西来测试终结点