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Its working fine now.Thank you very much.Actually i m new to frame work 2.0 and form tag.
"I am burning...the only thing rest in me is you..."
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glad to hear that.
Thanks and Regards,
Michael Sync ( Blog: http://michaelsync.net)
"Please vote to let me (and others) know if this answer helped you or not. A 5 vote tells people that your question has been answered successfully and that I've pitched it at just the right level. Thanks."
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This message means that you are trying to place a text box outside the page form. You need to ensure that you place the textbox between a form tag with runat="server". For example:
<br />
<form runat="server"><br />
<asp:TextBox id="textbox1" runat="server" /><br />
</form><br />
If you want the textbox to be in the table then ensure that the table is between the form tags.
The simplest solution is to add the form tag to the master page so that it wraps the content place holder, this means that any page that inherits the master page will automatically have the form tag. This is not best practice because it will mean that pages that do not use forms will have the form and therefore view state sent with them which increase page download times and eats your bandwidth, but for quick test solution its job done.
Hope this helps.
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Its working fine now.Thank you very much.Actually i m new to frame work 2.0 and form tag.
"I am burning...the only thing rest in me is you..."
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Hey guys
I have used OutPutCaching on a webpage. Now how I can measure what benifit I'm geting from Caching...!
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You can use the Application_BeginRequest() and Application_EndRequest() methods in the global.asax file to determine the time it took to process the request. But here your question is How to measure the Performance of Output Cache. You may have to take a look at this MSDN article[^], It says that performance of the Output Cache has to be evaluated based on the hits to the cached entries because Caching utilizes the server memory. If the hits are low to the Cached entries then sometime it may affect the performance.
- Regards - JON Life is not measured by the amount of breaths we take, but by the moments that take our breath away.
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Following is the chunk of my source file.There i was trying to add two form tags.in first tag(with runat server) i dropped linkbutton and in second form tag (without runat server), i dropeed textbox.Then i executed the page .
I got the error
<u><b>Control 'TextBox1' of type 'TextBox' must be placed inside a form tag with runat=server.</b></u>
<form id="form1" runat="server">
<div>
<asp:LinkButton ID="LinkButton1" runat="server">LinkButton</asp:LinkButton>
</div>
</form>
<form id="f1" >
<div>
<asp:TextBox ID="TextBox1" runat="server"></asp:TextBox>
</div>
</form>
-- modified at 7:38 Tuesday 20th November, 2007
Yesterday is a canceled check. Tomorrow is a promissory note. Today is the ready cash. USE IT.
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The second form is a normal plain vanilla HTML form. Hence it can not contain a serverside Textbox. You can opt for a simple Input Type textbox without a runat=server.
Vasudevan Deepak Kumar
Personal Homepage Tech Gossips
A pessimist sees only the dark side of the clouds, and mopes; a philosopher sees both sides, and shrugs; an optimist doesn't see the clouds at all - he's walking on them. --Leonard Louis Levinson
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Yeah I also got same error once, So if u can tell wot is ur requirement u can have any alternative...
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i think the code project site too contains a form tag which has search options(asp.net , vb.net , c# , database , discussions) to navigate from page to page ?
Yesterday is a canceled check. Tomorrow is a promissory note. Today is the ready cash. USE IT.
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can u see the categories mentioned above , named asp.net , c# , vb.net ...
i was talking about it.
Yesterday is a canceled check. Tomorrow is a promissory note. Today is the ready cash. USE IT.
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That I can very well, but those forms will not be having any control with runat="server" tag...
What is ur requirement?
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i was thimking , how to incorporate it in the real life project
Yesterday is a canceled check. Tomorrow is a promissory note. Today is the ready cash. USE IT.
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When I got the same problem, I used HTML controls and then complex javascript function....
The one thing (In my knowledge) which we can add with runat="server" is hidden variables..
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ok one more thing i wanted to ask. delegates i know is a safe type function pointer. y should i use it , i am not getting it's utility , the same thing can happen if i directly call the funciton , then what's the need to incorpoate the delagates?
Yesterday is a canceled check. Tomorrow is a promissory note. Today is the ready cash. USE IT.
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Dont have much idea abt it. You should paste it as a question, its a good one...
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Default, ASP.NET won't allow to have two server side form in one page. BTW, why do you want to do so ?
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it was asked to me in interview? i was blank face there.That's y asking
Yesterday is a canceled check. Tomorrow is a promissory note. Today is the ready cash. USE IT.
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Hi,
I have uploaded images to a db, there it was storing link filename(1.jpg), now i want to retrive those image to my new page, can u pls help me...
krishna.v
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In that example its storing as a link, but iam storing like file name, in db filed its storinglike 1.jpg, so i have to convert that from binary code,
i have tryed by using wht u have given, but i dint get a image.
krishna
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krishnavaradharajan wrote: but iam storing like file name,
then why not file path , that would be very easy for you.
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Retrieve the Image from the DataBase and set the Content Type of the Response Object and do a BinaryWrite of the ImageByte array using the Response Object . This Image display code should be on a seperate page. say if you want to display the Image in an image control then all you have to do is just set the ImageURL property of the control to the URL of this Image Display Page.
- Regards - JON Life is not measured by the amount of breaths we take, but by the moments that take our breath away.
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I have given the code like this.....
while (dr10.Read())
{
Response.Write("");
}
'uploads' is a folder, i have 1.jpg inside this folder and its uploaded in server, now i have 1.jpg in my db filed so i have retrive that 1.jpg in my new page. the which i was given is above.
krishna
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