gridview

Question

in gridview there are some linkbutton and butoon whose visible property is currently false...
i want it get visible when it meet some specicified conditions......
but my problem is how to set the property of these button visible=true....

Dipa Ahuja · Answer

in page_load event call one method like : showLink() in this method write the condition of link button like void showLink() { for (int i = 0; i LinkButton lk=(LinkButton)GridView1.Rows[i].FindControl("name of your link button"); / /condition to display if( ) { lk.Visible=true; } else { lk.Visible=false; } } } and also call this method in protected void GridView1_DataBound(object sender, EventArgs e) { showLink() }

Deepak Pandey · Answer

hello Chirag ... i tried your code it is not working......... when i tried to true the visible property of link button it throws an exception... plz try to give proper solution of my proble.....

Crish · Answer

hi in gridview there are one event RowDatabound. It uses for finding each and every row data bound in grid. You can find your linkbutton and button over there. In rowdatabound write below code Linkbutton lnkbutton = (Linkbutton)e.item.Findcontrol("linkbutton"); you can find this control value and you can set visible true or false accordingly as per your requirement. Don't forget to Mark Do you like this Answer that solved your problem!

Charith Liyanagamage · Answer

Hi Deepak, You can use below commands to enable select/edit/delete buttons of the gridview at any condition point. if(xxxxxxx) { //enable Select button GridView1.AutoGenerateSelectButton = true ; //enable Edit button GridView1.AutoGenerateEditButton = true ; //enable delete button GridView1.AutoGenerateDeleteButton = true ; }

gridview

Answers (4)