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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.
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// 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);
}
}
}
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Good Lord, tell me you didn't just post all the details of your paypal account to CP ?
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 )
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Kind of looks like it
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
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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
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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 ?
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 )
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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;
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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.
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 )
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Dear Christian Graus,
$get is a function in ASP.NET AJAX that it's functionality is the same document.getElementById
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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 )
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 )
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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
There is always something to learn
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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
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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.
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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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I think the problem occurs just like "Venkatesh Mookkan" said , in the post reply
if your problem similer to mine, I think the method solution by Venkatesh Mookkan should solve the preventing, but taking care for the access control identifier to proper according to your application,
actually I did not fully understand your problem , is it with button or with another control, is it from master page or something else,
clarify plz
There is always something to learn
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How to check if a data is to be entered using insert query statement,check the field we entered is muneric or not if numeric save it into database.Otherwise print a msg
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Use int.TryParse to see if a string contains a number.
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 )
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If a field datatype is represented as tinyint.By mistakenly if a user enter the data as a string value then an exception will be catched in sql.How to avoid the exception .Or if posiible to declare a msg inorder for this exception ..How i check the field as mumeric or not in code.
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SreejithAchutan wrote: By mistakenly if a user enter the data as a string value then an exception will be catched in sql.
You don't perform basic data validation like this in your database. You perform it in your web pages or in your middle tier. It is bad practice to send invalid input like this all the way to your database to be validated. Errors like this should be caught and handled far earlier in your application.
Paul Marfleet
"No, his mind is not for rent
To any God or government"
Tom Sawyer - Rush
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pmarfleet wrote: It is bad practice to send invalid input like this all the way to your database to be validated. Errors like this should be caught and handled far earlier in your application.
I agree. It is amazing how much bad practice I see around here sometimes.
"Real programmers just throw a bunch of 1s and 0s at the computer to see what sticks" - Pete O'Hanlon
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Hello Sir\Madam,
Please provide me info on how to link to the various forms in a project using coding but not by using properties window.
for e.g:
if(condn)
{
then to next form ;( pl provide code how to achieve this)
}
Thank you!
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Response.Redirect("http://yourdomain.com/nextpage.aspx")
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if(cndtn)<br />
{<br />
Response.Redirect("Next Form");<br />
}<br />
Erasers are for people who are willing to correct their mistakes.
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i have a data list and one delete button on it button after clicking delete button it will not fired can any one help me
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Can u post ur code...
Erasers are for people who are willing to correct their mistakes.
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