将 arrayList 从一个片段传递到另一个片段

Question

我想将一个 Arraylist（SearchModl 是一个类）从一个片段传递到另一个片段。我正在使用 Bundle 来执行此操作。但是，当我在其他片段中获取捆绑包时，我得到的是空值

搜索模型

public class SearchModel implements Serializable {

    private String code,category, minExp, maxExp, postedOn, counts, applied, position, desc, type, hour, status, expiryDate, address, gender, religion, summary, requestId, requestorId;

    public SearchModel(String code,String category, String minExp, String maxExp, String postedOn, String counts, String applied, String position, String desc, String type, String hour, String status, String expiryDate, String address, String gender, String religion, String summary, String requestId, String requestorId) {
        this.code=code;
        this.category = category;
        this.minExp = minExp;
        this.maxExp = maxExp;
        this.postedOn = postedOn;
        this.counts = counts;
        this.applied = applied;
        this.position = position;
        this.desc = desc;
        this.type = type;
        this.hour = hour;
        this.status = status;
        this.expiryDate = expiryDate;
        this.address = address;
        this.gender = gender;
        this.religion = religion;
        this.summary=summary;
        this.requestId = requestId;
        this.requestorId = requestorId;
    } 

片段 A

for (int i = 0; i < tableArray.length(); i++) {
                                        JSONObject table = tableArray.getJSONObject(i);
                                        data = new SearchModel(table.getString("Job_Code"), table.getString("Job_Category"), table.getString("Min_Exp"), table.getString("Max_Exp"), table.getString("Posted_On"), table.getString("Candidate_Counts"), table.getString("Applications"), table.getString("No_Of_Pos"), table.getString("Job_Desc"), table.getString("Job_Type"), table.getString("Job_Hours"), table.getString("Job_Status"), table.getString("Job_Exp_Date"), table.getString("Address"), table.getString("Gender_Name"), table.getString("Religion_Name"), table.getString("Exp_Summary"), table.getString("IJob_Request_ID"), table.getString("Requestor_Name"));
                                        values.add(data);

                                    }

//                                    getArrayList.getArray(values);


                                    bundle.putSerializable("array", (java.io.Serializable) values);

                                }
                            } catch (JSONException e1) {
                                e1.printStackTrace();
                            }
                        }
                        CommonFunctions.showProgress(getActivity(), "Please Wait...", false);
                        fragmentTransaction = getFragmentManager().beginTransaction();
                        SearchJobList searchJobList = new SearchJobList();
                        searchJobList.setArguments(bundle);
                        fragmentTransaction.replace(R.id.header_container, searchJobList, "searchJob").commit();
                    }
                }); 

第二个片段

public void onActivityCreated(Bundle savedInstanceState) {
        super.onActivityCreated(savedInstanceState);
        data = new ArrayList<SearchModel>();
        Bundle bundle = getArguments();
        data = (List<SearchModel>) bundle.getSerializable("array");
        getArrayList = new GetArrayList() {
            @Override
            public void getArray(List<SearchModel> listValues) {
                data = listValues;
                jobAdapter = new SearchJobAdapter(getActivity(), data);
                setListAdapter(jobAdapter);
            }
        };

    } 

Answer 1

您可以简单地将 ArrayList 保留在 Activity 中，供两个 Fragment 使用。

动态地，您在片段中创建一个公共方法，并在创建片段后，通过 setter 传递 List。

 YourActivity extends Activity {

   void createFragment(){
        YourFragment yourFragment = Fragment.getInstance();
        yourFragment.setList(models);
  }

}

YourFragment extends Fragment {
  private List<SearchModel> models; 

  public void setList(List<SearchModel> models){
        models = models;
  }
}

静态地，也不是最好的解决方案：

YourActivity extends Activity {
    static ArrayList<SearchModel> models;


}

YourFragment extends Fragment {

  public void doSomething(){
        List<SearchModel> model = YourActivity.models;
  }
}

实施 PArcelable 将是最好的解决方案：

http://developer.android.com/reference/android/os/Parcelable.html 这里有一个解决方案：Android ArrayList<MyObject> pass as parcelable

Answer 2

1.实现一个单例类作为传递对象的例子。 （希望你知道什么是单例，我不知道，添加评论告诉我。

class ExampleClass {
    private ExampleClass () {}

    static ExampleClass obj = nil;
    public ExampleClass instance() {
         if (obj == nil) obj = new ExampleClass ();
         return obj;
    }

    public ArrayList<SearchModel> cache;
 }

2.在from活动中，

ExampleClass .instance().cache = Cus_Obje_arraylist;

3.然后在to活动中，可以从bridge类中获取。

ArrayList<SearchModel> Cus_Obje_arraylist = ExampleClass.instance().cache;