Question Liaison de DataGrid à ObservableCollection


j'ai un ObservableCollection<Dictionary> et veulent le lier à un 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 que les clés deviennent l'en-tête du DataGrid et les valeurs de chaque Dictionary article à être les lignes. Réglage de la ItemsSource ne marche pas.


12
2018-01-05 10:59


origine


Réponses:


Vous pouvez utiliser un dictionnaire dynamique pouvant être lié. Cela exposera chaque entrée de 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));
    }
}

Ensuite, 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 a besoin d'un code personnalisé pour construire les colonnes:

XAML:

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

Événement AutoGeneratedColumns:

    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) });            
        }            
    }

23
2018-01-05 14:04



Sur la base de la réponse de Westons, je suis arrivé avec une autre solution sans utiliser une classe BindableDynamicDictionary personnalisée.

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

Il fait essentiellement la même chose que westons BindableDynamicDictionary avec l'inconvénient de ne pas avoir l'opérateur d'index disponible puisqu'il implémente explicitement 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) });
  }
}

Notez que la seule différence est que vous devez lancer le ExpandoObject à IDictionary<string, object> pour accéder / ajouter des valeurs ou des propriétés via l'opérateur d'index.


1
2017-11-24 10:05