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<customErrors defaultRedirect="ErrorPage.asp*x" mode="On">
<error statusCode="401" redirect="AccessDenied.aspx" />
</customErrors>
protected void Application_EndRequest(Object sender,
EventArgs e)
{
HttpContext context = HttpContext.Current;
if (context.Response.Status.Substring(0, 3).Equals("401"))
{
Response.Redirect("UnauthorisedPage.aspx");
}
}
if (!Request.IsAuthenticated)
{
Response.Redirect("UnauthorisedPage.aspx");
}
IIS > Virtual Directory properties > Custom Errors page > 401;2 > Properties
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I have a simple textbox with an initial text, when I type some other text in the textbox it and click my submit button, I see that the textbox.Text value stays the old initial value and the information which I typed has lost. How can I fix it?
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It sounds like the text is being re-initialised each postback. Try the following (in VB):
If not Page.IsPostback
// Initialise textbox
End If
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Have your verified EnableViewState is on for this control? Is the control being re-initialized during the postback?
I know the language. I've read a book. - _Madmatt
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what are the configuratiions in the asp.net validation controls? and please give the code to the configurations?
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Your question makes no sense. Please try to restate clearly what you a trying to do, what you have tried and what issues you have.
I know the language. I've read a book. - _Madmatt
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There is configurations. You mean properties?
Important(common) properties of validators are,
ControlToValidate , Display , ErrorMessage , EnableClientScript , IsValid , Text
Additional properties are,
CompareValidator - ControlToCompare, Operator, Type, ValueToCompare
CustomValidator - ClientValidationFunction, OnServerValidate
RangeValidator - MaximumValue, MinimumValue, Type
RegularExpressionValidator - ValidationExpression
RequiredFieldValidator -
ValidationSummary - DisplayMode, HeaderText, ShowMessageBox, ShowSummary
Read this
ASP.NET Validation Server Controls[^]
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I'm using msaccess(2007) database and I get the error message above while tryig to execute the following code:
OleDbCommand cmd = new OleDbCommand();
cmd.CommandText = "INSERT INTO tblCategory(CategoryName, OrderNum) " +
"VALUES(@CategoryName, @OrderNum) " +
"SELECT @@IDENTITY AS ID";
cmd.Parameters.Add(new OleDbParameter("@CategoryName", Category.CategoryName));
cmd.Parameters.Add(new OleDbParameter("@OrderNum", Category.Order));
cmd.Connection = DataAccess.CreateConnection("Malasot", false);
DataTable dt = new DataTable();
OleDbDataAdapter adapter = new OleDbDataAdapter(cmd);
if(cmd.Connection.State != ConnectionState.Open)
cmd.Connection.Open();
adapter.Fill(dt);
note: I have tryed putting semicolon between the 2 statements and it doesn't solve the problem.
modified on Monday, June 20, 2011 2:38 PM
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as your error stating, just place semi colon after the insert statement.
cmd.CommandText = "INSERT INTO tblCategory(CategoryName, OrderNum) " +
"VALUES(@CategoryName, @OrderNum); " +
"SELECT @@IDENTITY AS ID";
Parwej Ahamad
ahamad.parwej@gmail.com
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I've just tryed it, it didn't help
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How about adding a semicolon after the second SQL statement:
cmd.CommandText =
"INSERT INTO tblCategory(CategoryName, OrderNum) " +
"VALUES(@CategoryName, @OrderNum); " +
"SELECT @@IDENTITY AS ID;";
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Error is same OR giving different one? Can you try once to execute two separate statement on by one because I have double that Access generating instantly Identity value.
Thanks,
Parwej
Parwej Ahamad
ahamad.parwej@gmail.com
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I've tryed adding semicolon and the error message I get is: "Characters found after end of SQL statement."
I guess it will work well if I execute 2 seperated commands, but my goal is to execute all in one command, if it's possible.
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Hi Benames,
I haven't try from side, but I had searched on Google and they clearly mentioned that we can not execute multiple statement in single query against Ms Access DB. So may be cause of the error. So if possible execute two separated statement to achieve your goal.
Thanks,
Parwej
Parwej Ahamad
ahamad.parwej@gmail.com
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Ok, probably you are right.
It goes well when I run it in 2 different commands.
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guys;
i want to show an UpdateProgress when triggering an updatepanel to be updated. like this
<asp:button id="button1" runat="server" text="button1" />
<asp:upddatepanel id="upddatepanel1" runat="server">
<triggers>
<asp:asyncpostbacktrigger ControlID="button1" EventName="Click" />
</triggers>
</asp:upddatepanel>
<asp:UpdateProgress ID="ModalUpdateProgress1" DisplayAfter="0"
runat="server" AssociatedUpdatePanelID="upddatepanel1">
<ProgressTemplate>
Loading ...
</ProgressTemplate>
</asp:UpdateProgress>
what do you think?
Help people,so poeple can help you.
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You don't want to use UpdateProgress control?
Parwej Ahamad
ahamad.parwej@gmail.com
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Sir;
The existing UpdateProgress shows its content on asynchrnous postbaks just in the case of the postback control was a child of the AssociatedUpdatePanel if any.
what i am trhinking of is inheriting from UpdateProgress adn add a new property of type List<string> named AssociatedUpdatePanelTriggers that holds triggers ids of the AssociatedUpdatePanel and push them to the javascript version of the class on method Render() though ClientScript.RegisterStartupScript()
what do you think?
Help people,so poeple can help you.
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actually i worked with asp.net and now i will go throw MVC. i need example for MVC plz help me
thank u
j somasekhar
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Here you will find lot of usefull tutorials and videos. Check "Learning Resources" section. At the top of that section, there are also links for additionnal content.
http://www.asp.net/mvc[^]
Philippe Mori
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@Philippe Mori Good link.
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When I deploy my web application and when two persons (administrator, Encoder) logged in at same time to my web application then the pages gets interchanges between the logged in users.
We have develop the software in ASP.net VS-2008 and SQL Server 2005.
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Not nearly enough information to be able to help. e.g. does this happen on the dev box as well? Are there any exceptions reported? Is it your design? Also, not clear what you mean by 'pages get interchanges betwen th elogged in users'???
"If you think it's expensive to hire a professional to do the job, wait until you hire an amateur." Red Adair.
nils illegitimus carborundum
me, me, me
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you are developing a web application rite. there is no problem how many users will be logged in. suppose with same user name u will login with different browsers rite there will be nothing happens. first ask the question clearly i am not getting ur question
j somasekhar
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