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Click to See Complete Forum and Search --> : pop up the result after receiving the message from the servlet

lingmingfong
07-01-2003, 09:14 PM
I would like to know:

After user click a button on a page --> the page passed the action back to the servlet to do the performance instead on the page itself --> servlet passed back the result --> result shown on the screen.

normally we can use httpservletrequest or session to store the result information, these result will show on that page.

But I want to know is, is there a way for the page to know/detect when the result is passing back and activate a PopUp to inform the user about the result?

FYI, the page has many buttons, but only one button request this action.


Please help....thank.:rolleyes:
Jona
07-01-2003, 10:23 PM
Originally posted by lingmingfong
But I want to know is, is there a way for the page to know/detect when the result is passing back and activate a PopUp to inform the user about the result?

FYI, the page has many buttons, but only one button request this action.

What exactly do you mean? You mean to process a form server-side, and print the results in a popup window via Javascript?
About the button, it should work with the right button, as long as it is the submit button (<input type="submit">).

[J]ona
Khalid Ali
07-01-2003, 11:06 PM
You should be able to do that in th eservlet response,,,
create another page and in that pages onload event call a function that opens the opup window,whicl will display the results
lingmingfong
07-01-2003, 11:07 PM
Thank you all.

I found the solution, just activate the popup or alert in the <BODY>

:) so happy.
Khalid Ali
07-01-2003, 11:30 PM
You are welcome..:D