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Xamarin Android App Not Persist Any State (Session)
Rupesh Kahane
Nov 29 2016 9:03 AM
376
Reply
I have one simple layout1 which contains one EditText and Two button.
On first button click I am just saving the user text value using Shared Preferences.
and on second button click,showing another layout 2
.
I would like to maintain (Persist state) in Xamarin Android App from one layout to another layout.
I have used ISharedPreferences but it is not Persist any state.
Below is my code for layout1
public
class
MainActivity : Activity
{
protected
override
void
OnCreate(Bundle bundle)
{
ISharedPreferences sharedPrefernce;
base
.OnCreate(bundle);
SetContentView(Resource.Layout.Main);
Button button = FindViewById<Button>(Resource.Id.btnAdd);
button.Click +=
delegate
{
EditText nameBox = FindViewById<EditText>(Resource.Id.editText1);
string
name = nameBox.Text.ToString();
var lc = Application.Context.GetSharedPreferences(
"MyContacts"
, FileCreationMode.Private);
var cEdit = lc.Edit();
cEdit.PutString(
"Name"
, name);
cEdit.Apply();
Android.Widget.Toast.MakeText(
this
,
"Item Added"
, ToastLength.Short).Show();
nameBox.Text =
""
;
};
Button viewContact = FindViewById<Button>(Resource.Id.btnShowData);
viewContact.Click +=
delegate
{
SetContentView(Resource.Layout.ViewContactsActivity);
};
}
Below is my code for layout2
protected
override
void
OnCreate(Bundle savedInstanceState)
{
var textView3 = FindViewById<TextView>(Resource.Id.textView3);
try
{
var lc = Application.Context.GetSharedPreferences(
"MyContacts"
, FileCreationMode.Private);
string
retrivedName = lc.GetString(
"Name"
,
null
);
textView3.Text = retrivedName.ToString();
}
catch
(System.Exception ex)
{
textView3.Text = ex.Message.ToString();
}
}
Please suggest me How to persist a state in Xamarin Android App.
Thanks
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