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Binding an ObservableCollection to multiple listviews not working
Catch multiple exceptions at once?ListBox vs. ListView - how to choose for data bindingLazy load of images in ListViewDetermining Which ListViewItem Checkbox existed inDatabinding issue with stopwatched elapsedC# WPF - Listview Binding not workingBinding 2 ObservableCollections to ListviewChanges in ObservableCollection do not update ListViewWPF Listbox binding problemsElement added to ObservableCollection in ViewModel not displayed
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I'm new to programming and am having trouble figuring out how to get my ObservableCollection working on two different listviews across two different Windows in WPF. I think the problem is how I implemented my ObservableCollection, with the main culprit being this line here as shown in the SoundsWindow class:
ViewModel vm = this.DataContext as ViewModel;
This is how my listview displays my items added through:
vm.AddFile(fi.Name, file, 1);
The issue being that this does not update the other listview on the other window and neither is anything stored.
Heres my ViewModel:
public class ViewModel : INotifyPropertyChanged
public ViewModel()
SoundFiles = new ObservableCollection<SoundFile>();
public event PropertyChangedEventHandler PropertyChanged = delegate ;
private void OnPropertyChanged(string name)
PropertyChangedEventHandler handler = PropertyChanged;
handler(this, new PropertyChangedEventArgs(name));
public void AddFile(string fileName, string fileLocation, int groupID)
SoundFile soundfile = new SoundFile FileName = fileName, FileLocation = fileLocation, GroupID = groupID ;
SoundFiles.Add(soundfile);
private ObservableCollection<SoundFile> soundfiles;
public ObservableCollection<SoundFile> SoundFiles
get return soundfiles;
set
soundfiles = value;
OnPropertyChanged("SoundFiles");
SoundsWindow class:
public partial class SoundsWindow : Window
{
public SoundsWindow()
InitializeComponent();
this.DataContext = new ViewModel();
private void loadFileBtn1_Click(object sender, RoutedEventArgs e)
var ofd = new OpenFileDialog();
ofd.Multiselect = true;
ofd.Filter = "MP3 Files (*.mp3)
My SoundFile class:
public class SoundFile
public string FileName get; set;
public string FileLocation get; set;
public int GroupID get; set;
And the xaml:
<ListView x:Name="FilesList1" ItemsSource="Binding SoundFiles" HorizontalAlignment="Left" Height="379" Margin="10,35,0,0" VerticalAlignment="Top" Width="600">
<ListView.View>
<GridView>
<GridViewColumn Header="Name" DisplayMemberBinding="Binding Path=FileName" Width="230"/>
<GridViewColumn Header="Location" DisplayMemberBinding="Binding Path=FileLocation" Width="320"/>
<GridViewColumn Header="Group" DisplayMemberBinding="Binding Path=GroupID" Width="40"/>
</GridView>
</ListView.View>
</ListView>
For testing purposes the other window is a copy of SoundsWindow so that shouldnt be a problem.
As I pretty much have no idea what I'm doing iv checked about everything but am pretty sure the culprit is my implementation of how i add and store things in ObservableCollection. Any help would be immensely appreciated.
c# wpf listview mvvm observablecollection
asked Mar 28 at 8:37
zashen1zashen1
111 bronze badge
add a comment
|
I'm new to programming and am having trouble figuring out how to get my ObservableCollection working on two different listviews across two different Windows in WPF. I think the problem is how I implemented my ObservableCollection, with the main culprit being this line here as shown in the SoundsWindow class:
ViewModel vm = this.DataContext as ViewModel;
This is how my listview displays my items added through:
vm.AddFile(fi.Name, file, 1);
The issue being that this does not update the other listview on the other window and neither is anything stored.
Heres my ViewModel:
public class ViewModel : INotifyPropertyChanged
public ViewModel()
SoundFiles = new ObservableCollection<SoundFile>();
public event PropertyChangedEventHandler PropertyChanged = delegate ;
private void OnPropertyChanged(string name)
PropertyChangedEventHandler handler = PropertyChanged;
handler(this, new PropertyChangedEventArgs(name));
public void AddFile(string fileName, string fileLocation, int groupID)
SoundFile soundfile = new SoundFile FileName = fileName, FileLocation = fileLocation, GroupID = groupID ;
SoundFiles.Add(soundfile);
private ObservableCollection<SoundFile> soundfiles;
public ObservableCollection<SoundFile> SoundFiles
get return soundfiles;
set
soundfiles = value;
OnPropertyChanged("SoundFiles");
SoundsWindow class:
public partial class SoundsWindow : Window
{
public SoundsWindow()
InitializeComponent();
this.DataContext = new ViewModel();
private void loadFileBtn1_Click(object sender, RoutedEventArgs e)
var ofd = new OpenFileDialog();
ofd.Multiselect = true;
ofd.Filter = "MP3 Files (*.mp3)
My SoundFile class:
public class SoundFile
public string FileName get; set;
public string FileLocation get; set;
public int GroupID get; set;
And the xaml:
<ListView x:Name="FilesList1" ItemsSource="Binding SoundFiles" HorizontalAlignment="Left" Height="379" Margin="10,35,0,0" VerticalAlignment="Top" Width="600">
<ListView.View>
<GridView>
<GridViewColumn Header="Name" DisplayMemberBinding="Binding Path=FileName" Width="230"/>
<GridViewColumn Header="Location" DisplayMemberBinding="Binding Path=FileLocation" Width="320"/>
<GridViewColumn Header="Group" DisplayMemberBinding="Binding Path=GroupID" Width="40"/>
</GridView>
</ListView.View>
</ListView>
For testing purposes the other window is a copy of SoundsWindow so that shouldnt be a problem.
As I pretty much have no idea what I'm doing iv checked about everything but am pretty sure the culprit is my implementation of how i add and store things in ObservableCollection. Any help would be immensely appreciated.
c# wpf listview mvvm observablecollection
asked Mar 28 at 8:37
zashen1zashen1
111 bronze badge
4
Obviouslythis.DataContext = new ViewModel();is called in both windows. You do hence have two view model instances. Create a single one outside the Window constructor and pass it to the DataContext of both Windows.
– Clemens
Mar 28 at 8:40
Can we see the code you use to load the window please?
– Robin Bennett
Mar 28 at 8:42
Thanks for the reply, could you please show me how I would do that @Clemens . Thanks for the help.
– zashen1
Mar 28 at 8:46
1
One simple approach would be a static property in the ViewModel class, e.g.public static ViewModel Instance get; = new ViewModel();. Then assign the instance to a Window's DataContext byDataContext = ViewModel.Instance;
– Clemens
Mar 28 at 9:14
1
This is exactly what I needed. Thanks @Clemens you're a legend.
– zashen1
Mar 28 at 9:43
add a comment
|
I'm new to programming and am having trouble figuring out how to get my ObservableCollection working on two different listviews across two different Windows in WPF. I think the problem is how I implemented my ObservableCollection, with the main culprit being this line here as shown in the SoundsWindow class:
ViewModel vm = this.DataContext as ViewModel;
This is how my listview displays my items added through:
vm.AddFile(fi.Name, file, 1);
The issue being that this does not update the other listview on the other window and neither is anything stored.
Heres my ViewModel:
public class ViewModel : INotifyPropertyChanged
public ViewModel()
SoundFiles = new ObservableCollection<SoundFile>();
public event PropertyChangedEventHandler PropertyChanged = delegate ;
private void OnPropertyChanged(string name)
PropertyChangedEventHandler handler = PropertyChanged;
handler(this, new PropertyChangedEventArgs(name));
public void AddFile(string fileName, string fileLocation, int groupID)
SoundFile soundfile = new SoundFile FileName = fileName, FileLocation = fileLocation, GroupID = groupID ;
SoundFiles.Add(soundfile);
private ObservableCollection<SoundFile> soundfiles;
public ObservableCollection<SoundFile> SoundFiles
get return soundfiles;
set
soundfiles = value;
OnPropertyChanged("SoundFiles");
SoundsWindow class:
public partial class SoundsWindow : Window
{
public SoundsWindow()
InitializeComponent();
this.DataContext = new ViewModel();
private void loadFileBtn1_Click(object sender, RoutedEventArgs e)
var ofd = new OpenFileDialog();
ofd.Multiselect = true;
ofd.Filter = "MP3 Files (*.mp3)
My SoundFile class:
public class SoundFile
public string FileName get; set;
public string FileLocation get; set;
public int GroupID get; set;
And the xaml:
<ListView x:Name="FilesList1" ItemsSource="Binding SoundFiles" HorizontalAlignment="Left" Height="379" Margin="10,35,0,0" VerticalAlignment="Top" Width="600">
<ListView.View>
<GridView>
<GridViewColumn Header="Name" DisplayMemberBinding="Binding Path=FileName" Width="230"/>
<GridViewColumn Header="Location" DisplayMemberBinding="Binding Path=FileLocation" Width="320"/>
<GridViewColumn Header="Group" DisplayMemberBinding="Binding Path=GroupID" Width="40"/>
</GridView>
</ListView.View>
</ListView>
For testing purposes the other window is a copy of SoundsWindow so that shouldnt be a problem.
As I pretty much have no idea what I'm doing iv checked about everything but am pretty sure the culprit is my implementation of how i add and store things in ObservableCollection. Any help would be immensely appreciated.
c# wpf listview mvvm observablecollection
asked Mar 28 at 8:37
zashen1zashen1
111 bronze badge
I'm new to programming and am having trouble figuring out how to get my ObservableCollection working on two different listviews across two different Windows in WPF. I think the problem is how I implemented my ObservableCollection, with the main culprit being this line here as shown in the SoundsWindow class:
ViewModel vm = this.DataContext as ViewModel;
This is how my listview displays my items added through:
vm.AddFile(fi.Name, file, 1);
The issue being that this does not update the other listview on the other window and neither is anything stored.
Heres my ViewModel:
public class ViewModel : INotifyPropertyChanged
public ViewModel()
SoundFiles = new ObservableCollection<SoundFile>();
public event PropertyChangedEventHandler PropertyChanged = delegate ;
private void OnPropertyChanged(string name)
PropertyChangedEventHandler handler = PropertyChanged;
handler(this, new PropertyChangedEventArgs(name));
public void AddFile(string fileName, string fileLocation, int groupID)
SoundFile soundfile = new SoundFile FileName = fileName, FileLocation = fileLocation, GroupID = groupID ;
SoundFiles.Add(soundfile);
private ObservableCollection<SoundFile> soundfiles;
public ObservableCollection<SoundFile> SoundFiles
get return soundfiles;
set
soundfiles = value;
OnPropertyChanged("SoundFiles");
SoundsWindow class:
public partial class SoundsWindow : Window
{
public SoundsWindow()
InitializeComponent();
this.DataContext = new ViewModel();
private void loadFileBtn1_Click(object sender, RoutedEventArgs e)
var ofd = new OpenFileDialog();
ofd.Multiselect = true;
ofd.Filter = "MP3 Files (*.mp3)
My SoundFile class:
public class SoundFile
public string FileName get; set;
public string FileLocation get; set;
public int GroupID get; set;
And the xaml:
<ListView x:Name="FilesList1" ItemsSource="Binding SoundFiles" HorizontalAlignment="Left" Height="379" Margin="10,35,0,0" VerticalAlignment="Top" Width="600">
<ListView.View>
<GridView>
<GridViewColumn Header="Name" DisplayMemberBinding="Binding Path=FileName" Width="230"/>
<GridViewColumn Header="Location" DisplayMemberBinding="Binding Path=FileLocation" Width="320"/>
<GridViewColumn Header="Group" DisplayMemberBinding="Binding Path=GroupID" Width="40"/>
</GridView>
</ListView.View>
</ListView>
For testing purposes the other window is a copy of SoundsWindow so that shouldnt be a problem.
As I pretty much have no idea what I'm doing iv checked about everything but am pretty sure the culprit is my implementation of how i add and store things in ObservableCollection. Any help would be immensely appreciated.
c# wpf listview mvvm observablecollection
c# wpf listview mvvm observablecollection
asked Mar 28 at 8:37
zashen1zashen1
111 bronze badge
asked Mar 28 at 8:37
zashen1zashen1
111 bronze badge
asked Mar 28 at 8:37
zashen1zashen1
111 bronze badge
asked Mar 28 at 8:37
zashen1zashen1
111 bronze badge
asked Mar 28 at 8:37
zashen1zashen1
111 bronze badge
111 bronze badge
4
Obviouslythis.DataContext = new ViewModel();is called in both windows. You do hence have two view model instances. Create a single one outside the Window constructor and pass it to the DataContext of both Windows.
– Clemens
Mar 28 at 8:40
Can we see the code you use to load the window please?
– Robin Bennett
Mar 28 at 8:42
Thanks for the reply, could you please show me how I would do that @Clemens . Thanks for the help.
– zashen1
Mar 28 at 8:46
1
One simple approach would be a static property in the ViewModel class, e.g.public static ViewModel Instance get; = new ViewModel();. Then assign the instance to a Window's DataContext byDataContext = ViewModel.Instance;
– Clemens
Mar 28 at 9:14
1
This is exactly what I needed. Thanks @Clemens you're a legend.
– zashen1
Mar 28 at 9:43
add a comment
|
4
Obviouslythis.DataContext = new ViewModel();is called in both windows. You do hence have two view model instances. Create a single one outside the Window constructor and pass it to the DataContext of both Windows.
– Clemens
Mar 28 at 8:40
Can we see the code you use to load the window please?
– Robin Bennett
Mar 28 at 8:42
Thanks for the reply, could you please show me how I would do that @Clemens . Thanks for the help.
– zashen1
Mar 28 at 8:46
1
One simple approach would be a static property in the ViewModel class, e.g.public static ViewModel Instance get; = new ViewModel();. Then assign the instance to a Window's DataContext byDataContext = ViewModel.Instance;
– Clemens
Mar 28 at 9:14
1
This is exactly what I needed. Thanks @Clemens you're a legend.
– zashen1
Mar 28 at 9:43
4
4
Obviously
this.DataContext = new ViewModel(); is called in both windows. You do hence have two view model instances. Create a single one outside the Window constructor and pass it to the DataContext of both Windows.– Clemens
Mar 28 at 8:40
Obviously
this.DataContext = new ViewModel(); is called in both windows. You do hence have two view model instances. Create a single one outside the Window constructor and pass it to the DataContext of both Windows.– Clemens
Mar 28 at 8:40
Can we see the code you use to load the window please?
– Robin Bennett
Mar 28 at 8:42
Can we see the code you use to load the window please?
– Robin Bennett
Mar 28 at 8:42
Thanks for the reply, could you please show me how I would do that @Clemens . Thanks for the help.
– zashen1
Mar 28 at 8:46
Thanks for the reply, could you please show me how I would do that @Clemens . Thanks for the help.
– zashen1
Mar 28 at 8:46
1
1
One simple approach would be a static property in the ViewModel class, e.g.
public static ViewModel Instance get; = new ViewModel();. Then assign the instance to a Window's DataContext by DataContext = ViewModel.Instance;– Clemens
Mar 28 at 9:14
One simple approach would be a static property in the ViewModel class, e.g.
public static ViewModel Instance get; = new ViewModel();. Then assign the instance to a Window's DataContext by DataContext = ViewModel.Instance;– Clemens
Mar 28 at 9:14
1
1
This is exactly what I needed. Thanks @Clemens you're a legend.
– zashen1
Mar 28 at 9:43
This is exactly what I needed. Thanks @Clemens you're a legend.
– zashen1
Mar 28 at 9:43
add a comment
|
0
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4
Obviously
this.DataContext = new ViewModel();is called in both windows. You do hence have two view model instances. Create a single one outside the Window constructor and pass it to the DataContext of both Windows.– Clemens
Mar 28 at 8:40
Can we see the code you use to load the window please?
– Robin Bennett
Mar 28 at 8:42
Thanks for the reply, could you please show me how I would do that @Clemens . Thanks for the help.
– zashen1
Mar 28 at 8:46
1
One simple approach would be a static property in the ViewModel class, e.g.
public static ViewModel Instance get; = new ViewModel();. Then assign the instance to a Window's DataContext byDataContext = ViewModel.Instance;– Clemens
Mar 28 at 9:14
1
This is exactly what I needed. Thanks @Clemens you're a legend.
– zashen1
Mar 28 at 9:43