Angularjs 如何撤消JSOG.parse（）以返回API

设置

我有一个春季MVC

我有一个有角度的前端（按下按钮）从后端获取定义，将其名称更改为“NewName”，然后尝试用新名称将其放回

当前代码

  $scope.simpleSaveDefinition = function() {
    Definition.simpleGet({uuid:$scope.definition.uuid}, function (result) {
      result.name = "NewName";
      var stringResult = JSOG.stringify(result);
      Definition.update({'uuid':$scope.definition.uuid}, stringResult);
    });
  }

它正在呼叫： definition.service.js

        'simpleGet': {
            method: 'GET', isArray: false, 
            headers: {
                "Authorization": "ABC"
            }, url: "api/v1/definition/:uuid/deep",
            transformResponse: function (data) {
                return JSOG.parse(data);
            }
        },

        'update': {method: 'PUT', headers: {"Authorization": "ABC"},
            transformResponse: function (data) {
                return JSOG.parse(data);
            }
        },

definition.service.js

        'simpleGet': {
            method: 'GET', isArray: false, 
            headers: {
                "Authorization": "ABC"
            }, url: "api/v1/definition/:uuid/deep",
            transformResponse: function (data) {
                return JSOG.parse(data);
            }
        },

        'update': {method: 'PUT', headers: {"Authorization": "ABC"},
            transformResponse: function (data) {
                return JSOG.parse(data);
            }
        },

错误

Possibly unhandled rejection: {"data":{"timestamp":1529075556923,"status":400,"error":"Bad Request","message":"JSON parse error: Unresolved forward references for: ; nested exception is com.fasterxml.jackson.databind.deser.UnresolvedForwardReference: Unresolved forward references for: \n at [Source: (PushbackInputStream); line: 1, column: 5685]Object id [2] (for com.ibm.ispww.serviceregistry.domain.swagger.media.Schema) at [Source: (PushbackInputStream); line: 1, column: 3257], Object id [17] (for


    ...

(for com.myregistry.domain.assemblies.NamedObject) at [Source: (PushbackInputStream); line: 1, column: 5676].",
  "path" : "/api/v1/definition/fac4e40b-8b41-4eca-94d2-6054ee2a0bd7"
}

所以，错误似乎是

JSOG.parse(result) 
result.name = "NewNAME"
JSOG.stringify(result)

不会产生可接受的对象

有趣的是，如果我删除这些步骤，只获取对象（从不解析）并返回它，它就会工作

问题

---

是否有某种方法可以解析结果，编辑结果，然后将其解析并返回？似乎JSOG并没有按照我预期的方式工作。

Typo:

JSON.parse

；不

JSOG

。不需要手动解析JSON；$http服务会自动执行此操作。我想我是在用这个，但是使用JSON是有效的。谢谢@georgeawg