ASP.NET Discussion Boards

Re: [Membership And Roles] Check if user exists

Bursh.24-Dec-07 5:18

Bursh.

24-Dec-07 5:18

I need to know how to check that the user exists! Not what tables you have.

Don't you just hate coming up with something to put here?

Re: [Membership And Roles] Check if user exists

Paul Conrad24-Dec-07 5:23

Paul Conrad

24-Dec-07 5:23

Bursh. wrote:
Not what tables you have.

Huh? You have to have the users stored somewheres in a table. You query the username/password and if it returns with nothing, then user doesn't exist.

"The clue train passed his station without stopping." - John Simmons / outlaw programmer

Re: [Membership And Roles] Check if user exists

Bursh.24-Dec-07 11:09

Bursh.

24-Dec-07 11:09

I do. I have them stored in the tables created by the asp_regsql tool. I need to know how to query it and check if it returns null

Don't you just hate coming up with something to put here?

how to call a button click event of a parent window in child window

jagan12322-Dec-07 21:57

jagan123

22-Dec-07 21:57

by clicking one button in parent window the popup is opened and now on clicking on the popup button the parent window another button event has to call how could be possible.

Re: how to call a button click event of a parent window in child window

Bursh.23-Dec-07 3:36

Bursh.

23-Dec-07 3:36

If you're writing in VB you could do something like this in your on click sub:

<br />
Dim intActionNo As Integer = 0<br />
<br />
Select intActioNo<br />
    Case 0<br />
        'Do first action<br />
        intActionNo = 1<br />
    Case 1<br />
        'Do second action<br />
        intActionNo = 2 'Or alternatively revert to 0<br />
End Select<br />

Hope that helps

Don't you just hate coming up with something to put here?

Re: how to call a button click event of a parent window in child window

Michael Sync23-Dec-07 14:44

Michael Sync

23-Dec-07 14:44

You need to use Javascript for that. You can access the opener object. then, get the object of DOM and then, fire the click event.

For example ~

window.opener.document.getElementById("yourbtn").click();

Just an example. there might be typo error so you need to find the correct syntax based on this concept.

Thanks and Regards,
Michael Sync ( Blog: http://michaelsync.net)

how to show a hidden column values in the grid view in a tool tip

jagan12322-Dec-07 19:10

jagan123

22-Dec-07 19:10

iam having a grid view with three columns where the last coumn is hidden i want this column value must be show in a tool tip when the mouse goes to the gridview second column

Re: how to show a hidden column values in the grid view in a tool tip

Christian Graus22-Dec-07 19:40

Christian Graus

22-Dec-07 19:40

So what you need to do is define the text in a label, and add the text you want to show, as a tooltip on that label.

Christian Graus - Microsoft MVP - C++

"also I don't think "TranslateOneToTwoBillion OneHundredAndFortySevenMillion FourHundredAndEightyThreeThousand SixHundredAndFortySeven()" is a very good choice for a function name" - SpacixOne ( offering help to someone who really needed it ) ( spaces added for the benefit of people running at < 1280x1024 )

Enabling other controls on the page on doubleclick of a row in the GirdView control

Janu_M22-Dec-07 17:14

Janu_M

22-Dec-07 17:14

Hi,

I am developing a webproject in C#.net using .net2005.

My project contains two pages. First page displays data from the database in a datagrid view. when i double clicks a row in the datagrid it directs to the second page.

for this i am writing the following code in the rowcreated event of the datagrid.

e.Row.Attributes.Add("ondblclick", String.Format("window.location='" + Columns[i].ToString() + "'"));

here the arraylist Columns contains the following sample data

Columns.Add( "frmOtherPage.aspx?ListType=Empty&ViewType=Invoice&UniqueId=" + uniqueid.ToString());

In the datagrid i am not displaying this unique id.

on double click of the row i am getting the uniqueid of that particular row in the second page.

In second page i am displaying the data of the selected row in detail and diplaying the sub activities of this row in a data grid. In this grid also i am not displaying the activityid.

Here i have to display some controls to create a new sub activity. I am placing all these controls in a panel and displaying (.visible = true) them when user clicks the create sub activity button.

Here what i want is, on doubleclicking a row on the sub activities grid the panel for creating activity must be visible and i have to display the data of that row in those controls of the panel.

Please Help me. Its urgent.

Thanks in Advance

Ramu

How to make photos "appear clickable" ?

scm22ri22-Dec-07 10:46

scm22ri

22-Dec-07 10:46

Hi Everyone,

(I apologize for my terrible terminology)

Do you know when you "wave" your cursor over a photo or a url link and it appears as I would like to put it "clickable" or the cursor changes into a "hand". My issue is this, on my site I have a photo that I've changed the source code to and the photo is actually "clickable" but if you "wave" your cursor over the photo it dosen't appear "clickable" Does anyone know how I would fix that ? The source code is below.

<asp:hyperlink id="HyperLink1" runat="server" imageurl="~/Ringtone-Carrier-Photos/alltel-Complimentary Ringtones.jpg">

onClick="makeUrl('http://fbgdc.com/click/?s=8785&c=53726&subid=yahoo=flycell=alltel')"

Target="_blank"Alltell 15 Complimentary Ringtones</asp:hyperlink>

http://tinyurl.com/2tca8w (This is the page, just "wave" your cursor over the "alltell photo" and you will know what I'm talking about.)

Thanks everyone for there help.

Re: How to make photos "appear clickable" ?

Christian Graus22-Dec-07 10:53

Christian Graus

22-Dec-07 10:53

Handle the onmouseover and onmouseout events and change the cursor accordingly.

Re: How to make photos "appear clickable" ?

Venkatesh Mookkan23-Dec-07 15:39

Venkatesh Mookkan

23-Dec-07 15:39

You can use either ways:
* Use CssClass to set the Image's cursor:pointer (Use StyleSheet here)
or
* HyperLink1.Style.Add("cursor","pointer") in the code behind.

Regards,
Venkatesh Mookkan.
Software Engineer, India
My: Website | Yahoo Group | Blog Spot

paypal ver iam wrong

Mogaambo22-Dec-07 6:03

Mogaambo

22-Dec-07 6:03

// iam testing in sandbox enviroment and it is shows the following error

//"ErrorCode=10501&Desc=Invalid Configuration
//&Desc2=This transaction cannot be processed due to an invalid merchant configuration."
// ver iam wrong i enter all the correct details

using System;
using System.Data;
using System.Configuration;
using System.Collections;
using System.Web;
using System.Web.Security;
using System.Web.UI;
using System.Web.UI.WebControls;
using System.Web.UI.WebControls.WebParts;
using System.Web.UI.HtmlControls;
using com.paypal.sdk.services;
using com.paypal.sdk.profiles;
using com.paypal.sdk.core;
using com.paypal.sdk;
using com.paypal.sdk.util;
using System.ComponentModel;
using System.Web.SessionState;
public partial class paymantapi : System.Web.UI.Page
{
protected void Page_Load(object sender, EventArgs e)
{

}
protected void Button1_Click(object sender, EventArgs e)
{
com.paypal.sdk.services.NVPCallerServices caller = new com.paypal.sdk.services.NVPCallerServices();
IAPIProfile profile = ProfileFactory.createSignatureAPIProfile();
profile.APIUsername = "seller_1197901374_biz_api1.rediffmail.com";
profile.APIPassword = "1197901386";
profile.Subject = string.Empty;
profile.APISignature = "AJPRAFDEADN4KU5-uQOlH6F4lndTAo8rfA-5axq.xADfGzVZNCtI.U-A";
caller.APIProfile = profile;
NVPCodec encoder = new NVPCodec();
encoder["METHOD"] = "DoDirectPayment";
encoder["PAYMENTACTION"] = "sale";
encoder["AMT"] = "59.45";
encoder["CREDITCARDTYPE"] = "VISA";
encoder["ACCT"] = "4356645484999804";
encoder["EXPDATE"] = "1/2026";
encoder["CVV2"] = "125";
encoder["FIRSTNAME"] = "Test";
encoder["LASTNAME"] = "User";
encoder["STREET"] = "1 Main St";
encoder["CITY"] = "San Jose";
encoder["STATE"] = "CA";
encoder["ZIP"] = "95131";
encoder["COUNTRYCODE"] = "US";
encoder["CURRENCYCODE"] ="USD" ;
encoder["NOTIFYURL"] = "www.paypal.yashsoftech.com/ipn.aspx";
string pStrrequestforNvp = encoder.Encode();
string pStresponsenvp = caller.Call(pStrrequestforNvp);

NVPCodec decoder = new NVPCodec();
decoder.Decode(pStresponsenvp);
Response.Write(pStresponsenvp);
string strAck = decoder["ACK"];
if (strAck != null && (strAck == "Success" || strAck == "SuccessWithWarning"))
{
string pStrResQue = "TRANSACTIONID=" + decoder["TRANSACTIONID"] + "&" +
"AMT=" + decoder["AMT"] + "&" +
"AVSCODE=" + decoder["AVSCODE"] + "&" +
"CVV2MATCH=" + decoder["CVV2MATCH"];
Response.Redirect("DoDirectPaymentReceipt.aspx?"+pStrResQue);

}
else
{
string pStrError = "ErrorCode=" + decoder["L_ERRORCODE0"] + "&" +
"Desc=" + decoder["L_SHORTMESSAGE0"] + "&" +
"Desc2=" + decoder["L_LONGMESSAGE0"];

Response.Redirect("APIError.aspx?"+pStrError);

}
}
}

Re: paypal ver iam wrong

Christian Graus22-Dec-07 9:23

Christian Graus

22-Dec-07 9:23

Good Lord, tell me you didn't just post all the details of your paypal account to CP ?

Re: paypal ver iam wrong

Paul Conrad22-Dec-07 9:33

Paul Conrad

22-Dec-07 9:33

Kind of looks like it Eek! | :eek:

"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon

Re: paypal ver iam wrong [modified]

Mogaambo22-Dec-07 20:27

Mogaambo

22-Dec-07 20:27

I posted all the details , the details are those which i collected from books and also from codeproject, just type the paypal in search box u will get the code from ajith r nair article.

Actually the credit card which iam giving is a sandbox generated credit card.

my developer.paypal.com, username="megha_3983@rediffmail.com password=chandra550

my seller password = 197901323
my buyer password = 197901255
plz tell me ver iam missin the details

hope u reply me soon

modified on Sunday, December 23, 2007 4:06:07 AM

Re: paypal ver iam wrong

Christian Graus22-Dec-07 22:34

Christian Graus

22-Dec-07 22:34

chanu007 wrote:
Actually the credit card which iam giving is a sandbox generated credit card.

I know, it was the paypal account details, that you reiterate here, that left me incredulous.

Have you tried asking paypal ?

problem with javascript

morteza5722-Dec-07 5:43

morteza57

22-Dec-07 5:43

Hi, i used from this javascript code in my website an it works correctly by IE.
But it dos not work with Firefox.
what is it's equivalent compatible with Firefox?
Thank you.

$get('Panel1').style.left=100;
$get('Panel1').style.top =100;
$get('Panel1').style.posWidth=100;
$get('Panel1').style.posHeight=100;

Re: problem with javascript

Christian Graus22-Dec-07 9:25

Christian Graus

22-Dec-07 9:25

What is $get ? I've never seen that. document.getElementById is how you find an element, and you should do it once, check the returned value, then use that variable.

Re: problem with javascript

morteza5722-Dec-07 9:54

morteza57

22-Dec-07 9:54

Dear Christian Graus,
$get is a function in ASP.NET AJAX that it's functionality is the same document.getElementById

Re: problem with javascript

Christian Graus22-Dec-07 10:03

Christian Graus

22-Dec-07 10:03

OK, but it doesn't use AJAX, right ? SO, I'd just do it properly as per what I outlined. I'd also do what I suggested to see if it fixes FF, and if it doesn't, you can use Firebug in FF to debug your code and work out what the property is ( but what you have looks right to me )

Enable linkButton form outer page

Knowledgestudent22-Dec-07 0:48

Knowledgestudent

22-Dec-07 0:48

hi there,
I am designing ASP.net over VB
I have a linkButton in the master page and it is disabled, in the login page the login Button must enable that linkButton
when I use it immediately like this:

stdMaster.ChangeModeBtn.enable(True)
it does not work , it gives me something about the protected withEvents "is not accessible in this context because it is protected"

I know that I am missing something but I do not know what is it
please tell me

urgent help please Hmmm | :|

There is always something to learn

Re: Enable linkButton form outer page

Paul Conrad22-Dec-07 9:29

Paul Conrad

22-Dec-07 9:29

Knowledgestudent wrote:
urgent

No. Must have patience.

"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon

Re: Enable linkButton form outer page

Venkatesh Mookkan23-Dec-07 16:01

Venkatesh Mookkan

23-Dec-07 16:01

Knowledgestudent wrote:
stdMaster.ChangeModeBtn.enable(True)
it does not work , it gives me something about the protected withEvents "is not accessible in this context because it is protected"

ChangeModeBtn control, you are using is not a public control to access from the child page (Login). To access that control add a public subroutine or function in the Master Page which would enable or disable your control, like:

In MasterPage:

Public Sub EnableLink(value As Boolean)
    ChangeModeBtn.Enabled = value
End Sub

Access the Link from your child page like:

stdMaster.EnableLink(True)

I hope this would help you.

Regards,
Venkatesh Mookkan.
Software Engineer, India
My: Website | Yahoo Group | Blog Spot

ASP.Net "is not accessible in this context because it is Protected"

JEP_Dude10-Feb-09 16:16

JEP_Dude

10-Feb-09 16:16

I also have this problem. My research has shown that is is caused by ASP.Net opening two files and copying what's typed in one form into the other automatically. Yet when you attempt to reference a control from one form it states the above error.

My findings revealed that this problem has been around for sometime. It seems like there should be one simple correction to prevent this type of error from ever occurring. But thats just a dream.

I like to think that I understand the major parts of what causes the error. Yet the bigger question still remains ... how do I fix it?

Please help ... Thanks in advance.

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