ASP.NET & C# - Two values in a listbox?

Question

I'm adding ListItems to a ListBox from two controls, both are DropDownLists.

The ListItem has the properties ListItem.SelectedItem and ListItem.SelectedValue, but I also want the ListBox to keep track of which DropDownList the ListItem came from.

What would be the best way to do this?

Erix · Accepted Answer · 2010-12-23 13:51:10Z

5

You could set the ID of the dropdown programatically as a attribute of the listitem.

ListItem i = new ListItem();
i.Attributes["IDofDropDown"] = "SomeID";

answered Dec 23, 2010 at 13:51

Erix

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3 Comments

Canavar Over a year ago

I'm not sure, are these attributes transferred between postbacks ?

rtpHarry Over a year ago

Wouldn't this emit it in the html markup and make the code invalid?

Since_2008 Over a year ago

@SP, I tried what you suggested via lstTrainers.Items[number].Attributes["TrainerType"] = "Trainer"; lstTrainers.Items[number].Attributes.Add("TrainerType", ExTrainer"); But I'm getting an attribute count of 0 when I look at the ListItem. I'm passing the ListBox to a different object.

rtpHarry · Accepted Answer · 2010-12-23 13:49:38Z

0

Encode the .Value section so that its something like DropDownList1:SomeValue and then you can use String.Split to extract the info back out of .SelectedValue?

answered Dec 23, 2010 at 13:49

rtpHarry

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Comments

Canavar · Accepted Answer · 2010-12-23 13:50:24Z

0

ListItem item contains Text and Value properties that mapped from option element in HTML. So you can not add any other value to this structure. But maybe you can save the source dropdown's id to the Value of the ListItem.

answered Dec 23, 2010 at 13:50

Canavar

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Comments

StefanE · Accepted Answer · 2010-12-23 13:57:39Z

0

You could create an in memory dataset to store all the data needed and then bind the dataset to your ListBox. When ever a value is selected you will have access to more fields in code behind.

answered Dec 23, 2010 at 13:57

StefanE

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Comments

Since_2008 · Accepted Answer · 2011-05-19 13:24:37Z

0

The way I went about solving this, was creating an ArrayList, with another array inside for each item.

answered May 19, 2011 at 13:24

Since_2008

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