0
我已經通過很多主題進行了搜索，也找到了一些相關的答案，但是我仍然無法找到解決方案，因此我發佈了這個問題。在多個JSON字符串中查找一個節點的平均值
問題描述
EmployeeResponse1 = [{"Ques":"1","Rating":"7"},{"Ques":"2","Rating":"1"},{"Ques":"3","Rating":"6"},{"Ques":"4","Rating":"1"},{"Ques":"5","Rating":"1"},{"Ques":"6","Rating":"1"},{"Ques":"7","Rating":"7"},{"Ques":"8","Rating":"1"},{"Ques":"9","Rating":"1"},{"Ques":"10","Rating":"1"},{"Ques":"11","Rating":"1"},{"Ques":"12","Rating":"1"},{"Ques":"13","Rating":"1"},{"Ques":"14","Rating":"1"},{"Ques":"15","Rating":"1"},{"Ques":"16","Rating":"10"}]
EmployeeResponse2 = [{"Ques":"1","Rating":"5"},{"Ques":"2","Rating":"4"},{"Ques":"3","Rating":"7"},{"Ques":"4","Rating":"8"},{"Ques":"5","Rating":"5"},{"Ques":"6","Rating":"9"},{"Ques":"7","Rating":"10"},{"Ques":"8","Rating":"4"},{"Ques":"9","Rating":"9"},{"Ques":"10","Rating":"6"},{"Ques":"11","Rating":"6"},{"Ques":"12","Rating":"6"},{"Ques":"13","Rating":"7"},{"Ques":"14","Rating":"7"},{"Ques":"15","Rating":"9"},{"Ques":"16","Rating":"8"}]
我在C＃這兩個JSON字符串（可以有更多的）。現在我想作最後的JSON字符串，這將是這樣的：
EmployeeResponseAvg = [{"Ques":"1","Rating":"6"},{"Ques":"2","Rating":"2.5"},{"Ques":"3","Rating":"6.5"},{"Ques":"4","Rating":"4.5"},{"Ques":"5","Rating":"3"},{"Ques":"6","Rating":"5"},{"Ques":"7","Rating":"8.5"},{"Ques":"8","Rating":"2.5"},....,{"Ques":"16", "Rating": "9"}]
像我希望Ques = 1的評級應該是評分（字符串1的問答= 1）和評分（字符串2的問答= 1）的平均值...類似的其他問題
即，類似於FINAL = [{QUES = 1，RATING =（Emp1（Rating.WHERE（QUES = 1），Emp2（Rating.WHERE（QUES = 1），.AVERAGE），........ ............}]
工作至今
模型 - > SurveyResponse.cs
public class SurveyResponse
{
public string Ques { get; set; }
public string Rating { get; set; }
}
public class ResponseDataCalls
{
public static SurveyResponse PutData(string t, string v)
{
SurveyResponse s = new SurveyResponse();
s.Ques = t;
s.Rating = v;
return s;
}
}
的WebAPI RevGroupChartController.cs
代碼的public class RevGroupChartController : ApiController
{
private hr_toolEntities _db = new hr_toolEntities();
public object Get(int cid, int gid)
{
spiderChart obj = new spiderChart();
var group_employees = (from ge in _db.hrt_group_employee
where ge.fk_group_id == gid
select ge.fk_employee_id).ToList();
List<string> EMP = new List<string>();
List<string> SUP = new List<string>();
List<SurveyResponse> EmpResponse = new List<SurveyResponse>();
List<SurveyResponse> SupResponse = new List<SurveyResponse>();
List<List<SurveyResponse>> tmpEMP = new List<List<SurveyResponse>>();
List<List<SurveyResponse>> tmpSUP = new List<List<SurveyResponse>>();
foreach(var emp in group_employees)
{
int eid = Convert.ToInt32(emp);
var Data = (from d in _db.hrt_cycle_response
join g in _db.hrt_cycle_groups on d.hrt_cycle.pk_cycle_id equals g.fk_cycle_id
where d.fk_cycle_id == cid && g.fk_group_id == gid && d.fk_employee_id == eid
select new
{
d.response_employee_answers,
d.response_supervisor_answers
}).First();
EMP.Add(Data.response_employee_answers);
SUP.Add(Data.response_supervisor_answers);
}
foreach(var e in EMP)
{
//tmpEMP = new JavaScriptSerializer().Deserialize<TEMP>(e);
var s = new JavaScriptSerializer();
List<SurveyResponse> em = s.Deserialize<List<SurveyResponse>>(e);
tmpEMP.Add(em);
}
foreach (var s in SUP)
{
//tmpSUP = new JavaScriptSerializer().Deserialize<TEMP>(s);
var e = new JavaScriptSerializer();
List<SurveyResponse> sp = e.Deserialize<List<SurveyResponse>>(s);
tmpSUP.Add(sp);
}
var empl = _db.hrt_questions.Select(x => new { x.question_name }).ToList();
List<int[]> Emprating = new List<int[]>();
//int avgRating;
int cnt = 0;
foreach(var item in tmpSUP)
{
int noofQ = item.Count;
int[] i = new int[noofQ];
for (int y = 0; y > tmpSUP.Count; y++)
{
i[y] = Convert.ToInt32(item[cnt].Rating);
}
Emprating.Add(i);
cnt++;
}
//obj.Employee = Data.response_employee_answers;
//obj.Supervisor = Data.response_supervisor_answers;
obj.ques = new List<object>();
for (int i = 0; i < empl.Count; i++)
{
obj.ques.Add(empl[i].question_name);
}
return obj;
}
public class TEMP
{
public List<SurveyResponse> data { get; set; }
}
}
說明
傳遞一個週期ID和組ID ... 每個組有超過1名僱員和每個員工具有主管 因此，如果說組ID 1023有2名員工。 現在我們有2名員工和2個主管 我們必須爲他們每個人一個JSON記錄
LIKE DB TABLE RESPONSE {fk_emp_id, fk_sup_id, cycle_id, emp_reponse(json), supervisor_response(json)}
所以我需要作一次員工JSON字符串（其中包含所有評分的平均值） ，一個JSON字符串對於主管（再次，無論是JSONs的平均值） 有可能是任何數量的員工，根據組大小 和每個員工總會有一個主管
總之我要像AA串：
FinalEmployeeResponse = [{'Ques': '1', 'Rating': 'R1'}, {'Ques': '2', 'Rating': 'R2'}, {'Ques': '3', 'Rating': 'R3'}, {'Ques': '4', 'Rating': 'R4'}, ........, {'Ques': '16', 'Rating': 'R16'}]
這裏，R1 = AVERAGE(Emp1json.Rating.WHERE('Ques' = 1), Emp2json.Rating.WHERE('Ques' = 1), .....) 和
R2 = AVERAGE(Emp1json.Rating.WHERE('Ques' = 2), Emp2json.Rating.WHERE('Ques' = 2), .....)
...等等....
期待您的答覆。
我是新的堆棧溢出，如果我錯過了某些東西，請詢問更多細節。
如果還有其他更簡單的方法，那也是可以接受的。 – vohrahul 2014-11-21 05:51:59