I want to use AngularJs in a my page to filter a list of records, i found this article very helpful. the problem is in the article, the JSON.serialize is applied on a list like this

 public static String getlstAccount() {
     return JSON.serialize([SELECT Id, Name, Type, Phone FROM Account limit 10]);

what i want is to send a wrapper class instead of a list and i do not know how to use this fonction to achieve this , can it be like that ?

public static String getWrapperList() {
         return JSON.serialize(WrapperList);

where WrapperList is wrapper class that i created.

any help ?

  • are you having any issue with this? IMO you can serialize this way too. – Mr.Frodo Jul 6 '16 at 9:21
  • I did not try it yet , but it sounds very simplistic to do it like that, i will try it now – Merabti iBRAHIM Jul 6 '16 at 12:55

You can use this code for wrapping your account list

public static string getWrappedAccountList() {
    //Assuming you wrapper class name is 'Wrapper'
    List<Wrapper> WrapperList = new List<Wrapper>();

    for(Account account : [SELECT Id, Name, Type, Phone FROM Account limit 10]) {
        Wrapper wrap = new Wrapper();
        wrap.name = account.name;
        wrap.Type = account.Type;
        wrap.Phone = account.Phone;


    return JSON.serialize(WrapperList);

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.