CODEWITHSUNDEEP
c#
Ham Burg
Ham Burg

Reputation: 79

Order objects List items

In my app I have array of objects. Each object contains List with urls. I want to display Lists item in that order:

First item from first objects list
First item from second objects list
First item from third objects list
Second item from first objects list
..
etc

Right now I am using foreach loop:

            foreach (Account acc in account)
            {
                    listBox1.Items.Add(acc.ShowData())
            }

And public method inside Account class to get items:

public string ShowData()
{          
        string singleItem = LinksArray.First();
        LinksArray.RemoveAt(0);
        return singleItem;               
}

It works, but I think there might be more elegant way to do this. Do you have any idea?

Upvotes: 0

Views: 60

Answers (3)

Knetic
Knetic

Reputation: 2127

A method named ShowData should never modify the data it is showing. Instead, you might be better served by something like this;

public IEnumerable<String> GetData()
{
    return LinksArray;
}

Which you can then use;

foreach(Account acc in accounts)
{
    foreach(String data in acc.GetData())
    {
        // add to listbox items
    }
}

This keeps a clear separation between your data and the way it's displayed and read by consumers of the class. It also, optionally, opens you up to using LINQ.

Upvotes: 0

Enigmativity
Enigmativity

Reputation: 117010

This works for me:

List<Uri>[] arrayOfListsOfUris = ...

IEnumerable<Uri> sorted =
    arrayOfListsOfUris
        .Select(xs => xs.Select((x, n) => new { x, n }))
        .Concat()
        .OrderBy(y => y.n)
        .Select(y => y.x);

foreach (Uri uri in sorted)
{
    //Do something with each Uri
}

Upvotes: 0

Jacob
Jacob

Reputation: 78850

Try flattening everything into an set of index/URL pairs, then order by index:

var orderedUrls = objects
    .SelectMany(o => o.Urls.Select((url, idx) => new { Index = idx, Url = url }))
    .OrderBy(indexedUrl => indexedUrl.Index)
    .Select(indexedUrl => indexedUrl.Url)

Upvotes: 2

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