Update datagrid button click

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If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. SqlClient Imports System. New 'This call is required by the Windows Form Designer. InitializeComponent 'Add any initialization after the InitializeComponent call End Sub 'Form overrides dispose to clean up the component list.

DataGrid Me. Button Me. DataSet Me. SqlCommand CType Me. DataGrid1, System. BeginInit CType Me. Collectives on Stack Overflow. Learn more. Asked 3 months ago. Active 3 months ago.

Viewed 33 times. Add-Type -AssemblyName System. ShowDialog Any help would be amazing!! Improve this question. Listor Listor 63 5 5 bronze badges. Add a Solution. Sergey Alexandrovich Kryukov 3-Jan pm. Kschuler 3-Jan pm. Can you post the code you are currently using to put the text into the cell?

It might help us understand the issue. It is what it is 3-Jan pm. In the code below I use a field from details view to hold the string, I'm using a gridview and details view on the same form, sorry its hard to explain. Click OpenFileDialog1.

FileName 'assign filename to FilepathTextBox. This field is derived from a details view 'that comes from the dataset. Text 'not important DatestampDateTimePicker. You don't seem to be referencing a DataGrid anywhere in that code. It looks like you are just assigning the value to a textbox The one you marked as "derived" from a details view, and I'm not sure what that means.

Could you explain some more? Also, you can click the "Improve Question" button and add this code and more information to your original question. That way people don't have to read all of the comments to find new info. There is also an option the have the display set to details You can also have gridview and details the same form that is how I'm working You must have some bit of code that starts with Dim FilePathTextBox as Textbox And then assigns it to the editing control of the datagridviewcell?

Or something similar? How are you doing that? Dave Kreskowiak 3-Jan pm. The problem is you are assigning test and an image to a textbox and picturebox, but nowhere in this code are you assigning this data to any database, datatable, dataset, or anyn other datasource.