Add multiple values in List

Question

is it possible to add below in list in c#?

Suraj Kumar · Answer

try this, //EmployeeId,EmployeeName,EmplCode List string []> EmpDtls = new List string []>(); string [] empDtl1 = new string [] { "101" , "A" , "EMP001" }; string [] empDtl2 = new string [] { "102" , "B" , "EMP002" }; string [] empDtl3 = new string [] { "103" , "C" , "EMP003" }; EmpDtls.Add(empDtl1); EmpDtls.Add(empDtl2); EmpDtls.Add(empDtl3); //Remember array element is accessed by name not index

Sreekanth Reddy · Answer

@Suraj Kumar Good thought... Will it work as per expected ?

Susanta Rout · Answer

create a class with properties EmployeeId,EmployeeName,EmplCode public class Employee { public int EmployeeId{get,set} public string EmployeeName{get,set} public string EmplCode{get,set} } List Emplst=new List (); EmployeeList emp=new EmployeeList(); emp. EmployeeId=123; emp. EmployeeName="Ram"; emp. EmplCode="abc" Emplst.Add(emp); using a class you can add but directly you add only one value if you want directly List Emplst=new List (); string EmployeeName="Ram"; Emplst.Add(EmployeeName);

Munish Kumar · Answer

Hope this would help.. Create a class Employee public class Employee { public string EmployeeID; public string EmployeeCode; public string EmployeeName; } and, then create list of this class containing objects of the same like below.. List emp = new List () { new Employee { EmployeeID = "1" , EmployeeCode = "101" , EmployeeName = "A" } }; Console.WriteLine(emp[0].EmployeeID); Console.WriteLine(emp[0].EmployeeCode); Console.WriteLine(emp[0].EmployeeName);

Sreekanth Reddy · Answer

It is not possible with List. & ofcourse we are having Dictionary or Hashtable where : one argument as KEY another arguament as VALUE But in your case its like being single object split as argument. For example in feature one more property added like EMPSALARY, Then your expectaion would be like another argument which leads to re-modify(Not a generic). Better try to utilize the list itself... List obj = new List Now access like below : foreach(var a in obj) { //Access the values... Not matter how many properties even in future... }

Add multiple values in List

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