Liaison de DataGrid à ObservableCollection & lt; Dictionary & gt;

J'ai un ObservableCollection<Dictionary> et souhaitez lier à une DataGrid.

ObservableDictionary<String,Object> NewRecord1 = new ObservableDictionary<string,object>();

Dictionary<String,Object> Record1 = new Dictionary<string,object>();
Record1.Add("FirstName", "FName1");
Record1.Add("LastName", "LName1");
Record1.Add("Age", "32");

DictRecords.Add(Record1);

Dictionary<String, Object> Record2 = new Dictionary<string, object>();
NewRecord2.Add("FirstName", "FName2");
NewRecord2.Add("LastName", "LName2");
NewRecord2.Add("Age", "42");

DictRecords.Add(Record2);

Je voulais les clés pour devenir l'en-tête de la DataGrid et les valeurs de chaque Dictionary point à la ligne. Réglage de la ItemsSource ne fonctionne pas.

source d'informationauteur Manoj

  1. 23

    Vous pouvez utiliser un bindable dictionnaire dynamique. Cela permettra d'exposer chaque entrée du dictionnaire en tant que propriété.

    ///<summary>
///Bindable dynamic dictionary.
///</summary>
public sealed class BindableDynamicDictionary : DynamicObject, INotifyPropertyChanged
{
///<summary>
///The internal dictionary.
///</summary>
private readonly Dictionary<string, object> _dictionary;
///<summary>
///Creates a new BindableDynamicDictionary with an empty internal dictionary.
///</summary>
public BindableDynamicDictionary()
{
_dictionary = new Dictionary<string, object>();
}
///<summary>
///Copies the contents of the given dictionary to initilize the internal dictionary.
///</summary>
///<param name="source"></param>
public BindableDynamicDictionary(IDictionary<string, object> source)
{
_dictionary = new Dictionary<string, object>(source);
}
///<summary>
///You can still use this as a dictionary.
///</summary>
///<param name="key"></param>
///<returns></returns>
public object this[string key]
{
get
{
return _dictionary[key];
}
set
{
_dictionary[key] = value;
RaisePropertyChanged(key);
}
}
///<summary>
///This allows you to get properties dynamically.
///</summary>
///<param name="binder"></param>
///<param name="result"></param>
///<returns></returns>
public override bool TryGetMember(GetMemberBinder binder, out object result)
{
return _dictionary.TryGetValue(binder.Name, out result);
}
///<summary>
///This allows you to set properties dynamically.
///</summary>
///<param name="binder"></param>
///<param name="value"></param>
///<returns></returns>
public override bool TrySetMember(SetMemberBinder binder, object value)
{
_dictionary[binder.Name] = value;
RaisePropertyChanged(binder.Name);
return true;
}
///<summary>
///This is used to list the current dynamic members.
///</summary>
///<returns></returns>
public override IEnumerable<string> GetDynamicMemberNames()
{
return _dictionary.Keys;
}
public event PropertyChangedEventHandler PropertyChanged;
private void RaisePropertyChanged(string propertyName)
{
var propChange = PropertyChanged;
if (propChange == null) return;
propChange(this, new PropertyChangedEventArgs(propertyName));
}
}

    Alors vous pouvez l'utiliser comme ceci:

        private void testButton1_Click(object sender, RoutedEventArgs e)
{
//Creating a dynamic dictionary.
var dd = new BindableDynamicDictionary();
//access like any dictionary
dd["Age"] = 32;
//or as a dynamic
dynamic person = dd;
//Adding new dynamic properties.  
//The TrySetMember method is called.
person.FirstName = "Alan";
person.LastName = "Evans";
//hacky for short example, should have a view model and use datacontext
var collection = new ObservableCollection<object>();
collection.Add(person);
dataGrid1.ItemsSource = collection;
}

    Datagrid besoins de code personnalisé pour la construction de colonnes:

    XAML:

    <DataGrid AutoGenerateColumns="True" Name="dataGrid1" AutoGeneratedColumns="dataGrid1_AutoGeneratedColumns" />

    AutoGeneratedColumns événement:

        private void dataGrid1_AutoGeneratedColumns(object sender, EventArgs e)
{
var dg = sender as DataGrid;
var first = dg.ItemsSource.Cast<object>().FirstOrDefault() as DynamicObject;
if (first == null) return;
var names = first.GetDynamicMemberNames();
foreach(var name in names)
{
dg.Columns.Add(new DataGridTextColumn { Header = name, Binding = new Binding(name) });            
}            
}
  2. 1

    Basé sur weston réponse, je suis venu avec une autre solution sans l'aide d'un custom BindableDynamicDictionary classe.

    Il y a une classe appelée ExpandoObject dans l'espace de noms System.Dynamic(qui est très utilisée dans ASP.NET).

    Il se fait de la même chose que weston BindableDynamicDictionary avec l'inconvénient de ne pas avoir l'index de l'opérateur disponible depuis explicitement implémente l'interface IDictionary<string, object>

    private void MyDataGrid_AutoGeneratedColumns(object sender, EventArgs e)
{
var dg = sender as DataGrid;
dg.Columns.Clear();
var first = dg.ItemsSource.Cast<object>().FirstOrDefault() as IDictionary<string, object>;
if (first == null) return;
var names = first.Keys;
foreach (var name in names)
{
dg.Columns.Add(new DataGridTextColumn { Header = name, Binding = new Binding(name) });
}
}

    Avis que la seule différence ici est que vous devez jeter le ExpandoObject à IDictionary<string, object> d'accès/ajouter des valeurs de propriétés ou via l'index de l'opérateur.