Sort ListBox accoring to filename if Items are filepaths

Sort ListBox accoring to filename if Items are filepaths

By : Hangqi Wu
Date : November 14 2020, 04:48 PM
this one helps. After a bit more research in a different direction, I have found a fairly straightforward solution here, which I haven't thought about before that doesn't use collections but rather an "old fashioned" way.
code :
private void SortFileList()
    int length = listBoxOpenFiles.Items.Count;
    string[] keys = new string[length];
    ListBoxItem[] items = new ListBoxItem[length];

    for(int i = 0; i < length; ++i)
        keys[i] = ((listBoxOpenFiles.Items[i] as ListBoxItem).Tag as MyClass).FileName;
        items[i] = listBoxOpenFiles.Items[i] as ListBoxItem;

    Array.Sort(keys, items);

    for (int i = 0; i < length; ++i)

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ListBox sort after items update, is Items.SortDescriptions binding INotifyPropertyChanged?

ListBox sort after items update, is Items.SortDescriptions binding INotifyPropertyChanged?

By : Łukasz Grzegorski
Date : March 29 2020, 07:55 AM
should help you out I keep this for future if anyone need it... Now it's run fluidly.
I finally fixed it in this way:
code :
        View = (ListCollectionView)CollectionViewSource.GetDefaultView(ContactData.ParseXML(doc));
        View.SortDescriptions.Add(new SortDescription("State", ListSortDirection.Ascending));
        View.SortDescriptions.Add(new SortDescription("FullName", ListSortDirection.Ascending));
        MyListBox.ItemsSource = View.SourceCollection;
        while (IsStatusUpdateRunnig)
            ContactData[] contacts = ((ContactData[])View.SourceCollection);//MyListBox.ItemsSource);
            XDocument doc = XDocument.Load("http://localhost/contact/xml/status.php");
            foreach (XElement node in doc.Descendants("update"))
                ContactData[] its = contacts.Where(i => i.UserID.ToString() == node.Element("UserID").Value).ToArray();
                if (its.Length > 0)
                    Dispatcher.Invoke(new Action(() => { View.EditItem(its[0]); }));
            Dispatcher.Invoke(new Action(() => { View.CommitEdit(); }));
How to sort Listbox Items?

How to sort Listbox Items?

By : Neena Paul
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , I was just wondering if there was a way to sort the items in a listbox e.g. alphabetically descending, ascending? Soo far this is my code which basically adds files into the listbox: , you can easily do by
code :
 Dim query As IEnumerable(Of SampleData) = _
            datasource.OrderBy(Function(SampleData) s.FileNameX )
Sort C# ListBox Items

Sort C# ListBox Items

By : Marek Reichenbach
Date : March 29 2020, 07:55 AM
this will help I'm using C# Adapter method, I add elements to a list and after that I call that list to fill a listbox, here is my button for do that: , Reverse the list and start counting untill you reach the end:
code :
//Reverse the list
//Add new items
for(int i = 0; i < customers.Count; i++) 
    listCustomers.Items.Add(string.Format("Costumer ID # {0} is: {1}", (i+1).ToString(), customersSaldo[i].Saldo);
Sort array of filepaths by filename

Sort array of filepaths by filename

By : Zola Zola Mille Euro
Date : October 14 2020, 06:10 PM
hop of those help? File is a comparable class, which by default sorts pathnames lexicographically. Just use ,
code :
sort an array of object accoring to the values

sort an array of object accoring to the values

By : Jabal Torres
Date : October 05 2020, 05:00 AM
hope this fix your issue You could get the entries, pick the first one and destructure it to key and value and take the same for the other object, then return the chained sorting values.
code :
var array = [{ v21: 'sad sdd' }, { aaa: 'sad sdd' }, { v11: 'r err rr' }, { hf32: 'erwwer fgh' }, { z3f2: 'agfrr vbbar' }, { t142: 'gggoog  anfa' }, { u23: 'fa handle err' }];

array.sort((a, b) => {
    var [keyA, valueA] = Object.entries(a)[0],
        [keyB, valueB] = Object.entries(b)[0];
    return valueA.localeCompare(valueB) || keyA.localeCompare(keyB);

.as-console-wrapper { max-height: 100% !important; top: 0; }
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