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janet #1
button click ... linking of 2 pages :-(((( --pls help!
hey hi...
this in in continuation to my other post. i am trying hard
and cannot find out whats the problem. i am getting
crazy...please asp experts ...do resolve the problem...its
something very simple which i am doign wrong.
wehn i clcik to the submit button ....its absolutely dead
nothing happens!!!!!!!!!!!!!!! why so :-(((((((
this is my select.asp script:
<html>
<head>
<title> monika </title>
</head>
<body>
<form method="post" action="submit.asp">
<table><tr><td bgcolor="SILVER">
<BIG> <B> <FONT COLOR=blue>
WELCOME 'STUDENT_NAME'
TODAY IS <%=DATE()%></b></td><p></p></tr>
</table><p></p>
<input type = "text" size ="50" maxlength=90 name
= "txtStory"><p></P>
<input type="button" value="SUBMIT" >
</form>
</body>
</html>
this is my submit.asp script:
<html>
<head>
<title>monika m</title>
</head>
<body>
<%
Dim strStoryContent
DIM RSA
DIM QUERY1
DIM ConnString
DIM Conn
response.write "comes in submit"
ConnString = "Driver={MySQL ODBC 3.51 Driver};"
ConnString = ConnString & "Port=3306; "
ConnString = ConnString & "DATABASE=test; "
strStoryContent = Request.form("txtStory")
If Request.form("SUBMIT") <> "" then
response.write "come1"
If Request("txtStory")<>"" then
response.write "come2"
Set Conn = Server.CreateObject("ADODB.Connection")
Conn.Open ConnString
set RSA = Server.CreateObject("ADODB.Recordset")
QUERY1 = "INSERT INTO student_content
(student_id,story_id,status_student,story_original _story_wo
rd_count) VALUES
('1','100','draft','"&strStoryContent&"',4)"
Set rsa = conn.Execute(QUERY1)
End If
ELSE
RESPONSE.WRITE ("SUBMIT IS A ZERO LENGTH STRING")
end if
%>
</body>
</html>
janet Guest
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Ken Schaefer #2 Re: button click ... linking of 2 pages :-(((( --pls help!
You need a "submit" input element.
<input type="submit" value="submit" name="submit"
*not*
<input type="button">
A type="button" does not submit the form.
Cheers
Ken
"janet" <knowgroup@hotmail.com> wrote in message
news:02c301c366ef$6eebf440$a001280a@phx.gbl...
: hey hi...
: this in in continuation to my other post. i am trying hard
: and cannot find out whats the problem. i am getting
: crazy...please asp experts ...do resolve the problem...its
: something very simple which i am doign wrong.
: wehn i clcik to the submit button ....its absolutely dead
: nothing happens!!!!!!!!!!!!!!! why so :-(((((((
:
: this is my select.asp script:
:
: <html>
: <head>
: <title> monika </title>
: </head>
: <body>
: <form method="post" action="submit.asp">
: <table><tr><td bgcolor="SILVER">
: <BIG> <B> <FONT COLOR=blue>
: WELCOME 'STUDENT_NAME'
:
: TODAY IS <%=DATE()%></b></td><p></p></tr>
: </table><p></p>
: <input type = "text" size ="50" maxlength=90 name
: = "txtStory"><p></P>
: <input type="button" value="SUBMIT" >
: </form>
: </body>
: </html>
:
: this is my submit.asp script:
: <html>
: <head>
: <title>monika m</title>
: </head>
: <body>
: <%
: Dim strStoryContent
: DIM RSA
: DIM QUERY1
: DIM ConnString
: DIM Conn
: response.write "comes in submit"
: ConnString = "Driver={MySQL ODBC 3.51 Driver};"
: ConnString = ConnString & "Port=3306; "
: ConnString = ConnString & "DATABASE=test; "
: strStoryContent = Request.form("txtStory")
:
: If Request.form("SUBMIT") <> "" then
: response.write "come1"
: If Request("txtStory")<>"" then
: response.write "come2"
: Set Conn = Server.CreateObject("ADODB.Connection")
: Conn.Open ConnString
: set RSA = Server.CreateObject("ADODB.Recordset")
: QUERY1 = "INSERT INTO student_content
: (student_id,story_id,status_student,story_original _story_wo
: rd_count) VALUES
: ('1','100','draft','"&strStoryContent&"',4)"
: Set rsa = conn.Execute(QUERY1)
: End If
: ELSE
: RESPONSE.WRITE ("SUBMIT IS A ZERO LENGTH STRING")
: end if
: %>
: </body>
: </html>
:
:
Ken Schaefer Guest
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janet #3 Re: button click ... linking of 2 pages :-(((( --pls help!
hey thanks buddy ... a sign of relief and i am on the
next page...
but still its goign to the else part. How can i use this
validation :
If Request.form("SUBMIT") <> "" then
----
this submit value is the one which i get when i click the
submit button. but still my code is not getting executed.
my 'if' condition is not getting satisfied???? why so? any
light?
hard>-----Original Message-----
>You need a "submit" input element.
>
><input type="submit" value="submit" name="submit"
>
>*not*
>
><input type="button">
>
>A type="button" does not submit the form.
>
>Cheers
>Ken
>
>"janet" <knowgroup@hotmail.com> wrote in message
>news:02c301c366ef$6eebf440$a001280a@phx.gbl...
>: hey hi...
>: this in in continuation to my other post. i am tryingproblem...its>: and cannot find out whats the problem. i am getting
>: crazy...please asp experts ...do resolve thedead>: something very simple which i am doign wrong.
>: wehn i clcik to the submit button ....its absolutely(student_id,story_id,status_student,story_original _story_wo>: nothing happens!!!!!!!!!!!!!!! why so :-(((((((
>:
>: this is my select.asp script:
>:
>: <html>
>: <head>
>: <title> monika </title>
>: </head>
>: <body>
>: <form method="post" action="submit.asp">
>: <table><tr><td bgcolor="SILVER">
>: <BIG> <B> <FONT COLOR=blue>
>: WELCOME 'STUDENT_NAME'
>:
>: TODAY IS <%=DATE()%></b></td><p></p></tr>
>: </table><p></p>
>: <input type = "text" size ="50" maxlength=90 name
>: = "txtStory"><p></P>
>: <input type="button" value="SUBMIT" >
>: </form>
>: </body>
>: </html>
>:
>: this is my submit.asp script:
>: <html>
>: <head>
>: <title>monika m</title>
>: </head>
>: <body>
>: <%
>: Dim strStoryContent
>: DIM RSA
>: DIM QUERY1
>: DIM ConnString
>: DIM Conn
>: response.write "comes in submit"
>: ConnString = "Driver={MySQL ODBC 3.51 Driver};"
>: ConnString = ConnString & "Port=3306; "
>: ConnString = ConnString & "DATABASE=test; "
>: strStoryContent = Request.form("txtStory")
>:
>: If Request.form("SUBMIT") <> "" then
>: response.write "come1"
>: If Request("txtStory")<>"" then
>: response.write "come2"
>: Set Conn = Server.CreateObject("ADODB.Connection")
>: Conn.Open ConnString
>: set RSA = Server.CreateObject("ADODB.Recordset")
>: QUERY1 = "INSERT INTO student_content
>:>: rd_count) VALUES
>: ('1','100','draft','"&strStoryContent&"',4)"
>: Set rsa = conn.Execute(QUERY1)
>: End If
>: ELSE
>: RESPONSE.WRITE ("SUBMIT IS A ZERO LENGTH STRING")
>: end if
>: %>
>: </body>
>: </html>
>:
>:
>
>
>.
>janet Guest
-
Ken Schaefer #4 Re: button click ... linking of 2 pages :-(((( --pls help!
Do this (on your second page), to see what is actually in your Request.Form
collection:
<%
For Each Item in Request.Form
Response.Write(Item & " = " & Request.Form(Item) & "<br>")
Next
%>
which should give you a listing of all the different items that have been
submitted, and their values.
Cheers
Ken
"janet" <knowgroup@hotmail.com> wrote in message
news:030e01c366f4$3c21fe60$a001280a@phx.gbl...
: hey thanks buddy ... a sign of relief and i am on the
: next page...
:
: but still its goign to the else part. How can i use this
: validation :
: If Request.form("SUBMIT") <> "" then
: ----
:
: this submit value is the one which i get when i click the
: submit button. but still my code is not getting executed.
: my 'if' condition is not getting satisfied???? why so? any
: light?
Ken Schaefer Guest
-
janet #5 WONDERFUL!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
hey ken..........
superb.........
u r were abs right...
i had to make the following changes:
If Request.form("send") = "SUBMIT" then
and its working.....
i wish i cud treat u ...;-)
thanks tremendously!
your Request.Form>-----Original Message-----
>Do this (on your second page), to see what is actually in& "<br>")>collection:
>
><%
>For Each Item in Request.Form
>
> Response.Write(Item & " = " & Request.Form(Item)items that have been>
>Next
>%>
>
>which should give you a listing of all the differentthe>submitted, and their values.
>
>Cheers
>Ken
>
>
>"janet" <knowgroup@hotmail.com> wrote in message
>news:030e01c366f4$3c21fe60$a001280a@phx.gbl...
>: hey thanks buddy ... a sign of relief and i am on the
>: next page...
>:
>: but still its goign to the else part. How can i use this
>: validation :
>: If Request.form("SUBMIT") <> "" then
>: ----
>:
>: this submit value is the one which i get when i clickexecuted.>: submit button. but still my code is not gettingany>: my 'if' condition is not getting satisfied???? why so?>: light?
>
>
>.
>janet Guest
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