This is a sample for ASP.Net to show, how easy it is to 'force' a default button for a textbox.

Question

User95761738 posted
Submitted by : Dinesh Velupula, WSDOT Matlab Certified Java2 Developer, Microsoft Certified Professional - SQL Server2000 C# & VB.NET Expert Deveoper ------------------------------------------------------------------------------------------------ This is a sample made for Microsoft ASP.Net to show, how easy it is to 'force' a default button for a textbox. So far you have needed to install various DLL's or controls to control what button must be clicked, when a user is submitting some data from an textbox by pressing [ENTER] key. Now this is a lot easyer using only a single line of code (when the my small function is included) to relate/assign the [ENTER]-key event from a textbox to the wanted button. The sub-routine is as followes: Public Sub DefaultButton(ByRef Page As System.Web.UI.Page, ByRef objTextControl As TextBox, ByRef objDefaultButton As Button) ' Sets default buttons. ' Created by Janus Kamp Hansen - http://www.kamp-hansen.dk Dim sScript As New System.Text.StringBuilder() sScript.Append("<script language=""javascript"">" & vbCrLf) sScript.Append("function fnTrapKD(btn){" & vbCrLf) sScript.Append(" if (document.all){" & vbCrLf) sScript.Append(" if (event.keyCode == 13)" & vbCrLf) sScript.Append(" { " & vbCrLf) sScript.Append(" event.returnValue=false;" & vbCrLf) sScript.Append(" event.cancel = true;" & vbCrLf) sScript.Append(" btn.click();" & vbCrLf) sScript.Append(" } " & vbCrLf) sScript.Append(" } " & vbCrLf) sScript.Append("}" & vbCrLf) sScript.Append("</script>" & vbCrLf) objTextControl.Attributes.Add("onkeydown", "fnTrapKD(document.all." & objDefaultButton.ID & ")") Page.RegisterStartupScript("ForceDefaultToScript", sScript.ToString) End Sub This is how you use the sub-function inside you own code: DefaultButton(Page, TextBox1, Button1) DefaultButton(Page, TextBox2, Button2) DefaultButton(Page, TextBox3, Button3)

Answer 1

User1222855324 posted
GREAT!

Answer 2

User-522307627 posted
GREAT! x2!!! I changed the Button to an ImageButton and changed the line: objTextControl.Attributes.Add("onkeydown", "fnTrapKD(document.all." + objDefaultButton.ID + ")"); to: objTextControl.Attributes.Add("onkeydown", "fnTrapKD(document.all." + objDefaultButton.ClientID + ")"); because it is inside of a usercontrol and it works great! Thanks

Answer 3

User564684961 posted
thank you guys very much! I have been searching all day for this solution, and it works!

Answer 4

User-376809674 posted
Awesome work bro. We been looking for something liek this forever. John

Answer 5

User-376809674 posted
I have one question.... since I don't know js for a darn, what is the sinifigance of the text box parameter. usually, when i submit a form I want all the text box info submitted. It appears, in this example, that the text box parameter doesn't really do anything. John

Answer 6

User2064163041 posted
I have copied the code above EXACTLY as it is into a user control but when I try and run it I get the following error message: Compiler Error Message: BC30648: String constants must end with a double quote. Source Error: Line 55: Line 56: sScript.Append("</script>" & vbCrLf) I can't see anything wrong in this line. Anyone any ideas?

Answer 7

User-912481521 posted
Take a look @ PRB: The </script> Tag Is Treated As a Closing Tag When Inside a Quoted String

Answer 8

User2064163041 posted
Thank you. I changed to line to: sScript.Append(chr(60) & "/script>" & vbCrLf) and now it works fine. Cheers.

Answer 9

User1928316552 posted
to Velupula: i have a question, do this event doesn't need any postback from the page... thanks... regards, dave

Answer 10

User1554345290 posted
I tried to use this script in conjunction with a user control, and received a javascript error when the key down event fired. The javascript function reported that the btn parameter was null or unrecognized. I changed the line: objTextControl.Attributes.Add("onkeydown", "fnTrapKD(document.all." & objDefaultButton.ID & ")") to read: objTextControl.Attributes.Add("onkeydown", "fnTrapKD(document.all." & objDefaultButton.ClientID & ")") Using ClientID appends the user control identifier and supplies the unique identification to the document that it needs when using user controls within a page.

Answer 11

User-207799750 posted
Super code! works like a charm with the ..ClientId change. Here's the C# version of the method. Don't forget to import access to StringBuilder, "using System.Text;" protected void DefaultButton(System.Web.UI.Page Page, System.Web.UI.WebControls.TextBox txtControl, System.Web.UI.WebControls.Button btnDefault) { //code reference:http://www.asp.net/Forums/ShowPost.aspx?tabindex=1&PostID=138589 //Created by Janus Kamp Hansen - http://www.kamp-hansen.dk StringBuilder sScript = new StringBuilder(); sScript.Append("<script language='javascript'>" + "\n"); sScript.Append("function fnTrapKD(btn){" + "\n"); sScript.Append(" if (document.all){" + "\n"); sScript.Append(" if (event.keyCode == 13)" + "\n"); sScript.Append(" { " + "\n"); sScript.Append(" event.returnValue=false;" + "\n"); sScript.Append(" event.cancel = true;" + "\n"); sScript.Append(" btn.click();" + "\n"); sScript.Append(" } " + "\n"); sScript.Append(" } " + "\n"); sScript.Append("}" + "\n"); sScript.Append("</script>"); txtControl.Attributes.Add("onkeydown", "fnTrapKD(document.all." + btnDefault.ClientID + ")"); Page.RegisterStartupScript("ForceDefaultToScript", sScript.ToString()); }

Answer 12

User913124150 posted
Great!!!

Answer 13

User-1449169821 posted
Andy Smith has a great control for this already...very reliable, Designer support http://www.metabuilders.com/Tools/DefaultButtons.aspx

Answer 14

User-973072541 posted
One problem I'm having with this is that if I want to use this code in an .ascx file, how do I know what the eventual id for the button is going to be?

Answer 15

User-973072541 posted
BTW does anyone know how to make this work on Mozilla?

Answer 16

User-323455760 posted
Hi, This code works very good for textboxes. But what to do if I have a textbox within a DataGrid? How to use the "DefautlButton(Page, ?????, ImageButton1)" function with a datagid? Thanks

Answer 17

User483808541 posted
TOOL WORKS GREAT! It took me a bit to figure out to but the DefaultButton(Page, txtPassword, lnkLogin) in the page_load procedure and keep it outside my if page.ispostback = false statement, example: Private Sub Page_Load(....) DefaultButton(Page, txtPassword, Button1) If Page.IsPostBack = False Then .... End If Exit Sub

Answer 18

User-635437160 posted
Thank u thank u thank u!!!!!!!

Answer 19

User-130293678 posted
Yeah, works as well with ImageButtons and LinkButtons... Keep up the good work!!!!!!

Answer 20

User690078794 posted
Thanks! This is a smart code. It works!

Answer 21

User-1917735787 posted
Doesn't seem to wirk with other browsers then IE

Answer 22

User-1696021186 posted
That function is cool, but I've experienced some problems with adding JavaScript like that to a page. To accomplish that task, I usually use a single line in Page_Load: Private Sub Page_Load(... ) Handles MyBase.Load Page.RegisterHiddenField("__EVENTTARGET", "cmdButton") End Sub

Answer 23

User-1628317575 posted
I had the same problem. Thank you!

Answer 24

User-46888941 posted
Nice code :)

Answer 25

User1019781410 posted
Has anyone got this working with browsers other than IE?

Answer 26

User1941284925 posted
i found similar code on this site:
http://www.beansoftware.com/ASP.NET-Tutorials/Accept-Enter-Key.aspx

again, it works in IE, but not in firefox.

it looks like 2.0 has some default button functionality, but the way they describe it won't work in my situation.

anyone have luck getting it to work elsewhere?

c

Answer 27

User1726897791 posted

Good point.

Property "ID" and "ClientID" are different some time, within a user control is a good example.

When the control is in a user control, the client side ID of the control will be prefixed with "UserControlName_", the nature of this solution is client side javascript, so got the ClientID is important.