---
title: "How to add listview items in C#"  
description: "How to add listview items in C#"  
author: "Manoj Bhatt"  
published: 2013-07-17  
updated: 2013-07-17  
canonical: https://www.mindstick.com/forum/1303/how-to-add-listview-items-in-c-sharp  
category: "c#"  
tags: ["c#"]  
reading_time: 1 minute  

---

# How to add listview items in C#

Hi [mindstick](https://yourviews.mindstick.com/view/84668/how-mindstick-is-used-for-marketing-purposes)!

Hello I have some doubts in listveiw I have a [listview](https://www.mindstick.com/articles/12744/listview-in-android) with two columns, i want to [add items](https://www.mindstick.com/forum/160381/how-to-add-items-to-a-list-collection-in-c-sharp) to the first [column](https://www.mindstick.com/forum/33860/how-to-calculate-column-summary-in-sql-server) which are given programmatically and the other column is from [database](https://www.mindstick.com/articles/12226/use-of-database-in-sencha-extjs-and-insert-record-from-user-form-using-ajax)... like

column1 column2

abc from database

def "

hij "

can this be done.... can any one help me out with this...

I am using C# ,[visual studio](https://www.mindstick.com/articles/12378/visual-studio-for-mac-is-out-of-beta-preview-now-officially-available)........

thanks in [advance](https://www.mindstick.com/blog/33258/jee-mains-and-jee-advance-exams)

## Replies

### Reply by shreesh chandra shukla

Solution!

I'm not sure how you want to match the programmatically added values and the db values, but you could create a new listitem for each of the database values i.e.

ListItem item = new ListItem("apple",[database value]);

[Add](https://www.mindstick.com/forum/12983/add-address-in-textbox-when-page-is-load-and-if-i-search-address-in-textbox-show-in-map-by-javascript) the [items](https://www.mindstick.com/forum/33812/getting-number-of-items-selected-in-uicollectionview-in-ios) to a collection

ItemCollection.Items.Add(item);

Then bind the collection to your control

ListView.DataSource = ItemCollection;


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