Hi,
This aspx page (let's call it thispage.aspx) fetches data from a
sqldatasource, then performs several things (in code-behind) and, to
simplify, passes data from code-behind via a hiddenfield to a javascript in
the aspx file. This javascript performs things and finally send data via a
form to another database.
My problem is that when the page is postback (with this java-line:
document.getElementById("ins").action="thispage.as px"), instead of
performing the code after "If Page.IsPostBack Then", it shows the original
aspx file again.
Any idea what i have to change to do what i want to do?
Thanks
Harry
aspx file:
-------
<form id="form1" runat="server">
<asp:SqlDataSource ID="SqlDataSource1" runat="server"
ConnectionString="Provider = ... ; Data Source =...;">
</asp:SqlDataSource>
<asp:HiddenField ID="HiddenField1" runat="server" />
</form>
<form id="ins" style="width:750px;background-color:Gray" method="post">
<input id="sql" name="sql" type="hidden" />
<input id="conn" name="conn" type="hidden" />
<input runat="server" id="Submit1" type="button" value="Klik hier om
te bewaren" onclick="sendtodb()"/>
</form>
<script language="javascript" type="text/javascript">
function sendtodb()
{
var nfieldout=document.getElementById("hiddenfield1"). value
......
......
document.getElementById("sql").value=inscomm
document.getElementById("conn").value=conn
document.getElementById("ins").action="thispage.as px"
document.getElementById("ins").submit()
return true;
</script>
code-behind
-----------
Protected Sub Page_Load(ByVal sender As Object, ByVal e As
System.EventArgs) Handles Me.Load
If Page.IsPostBack Then
Dim conn, sql
sql = Request.Form("sql")
conn = Request.Form("conn")
Dim oConnection As OleDbConnection
Dim sConnection As String
oConnection = New OleDbConnection()
Dim comd As OleDbCommand
sConnection = conn
oConnection.ConnectionString = sConnection
oConnection.Open()
comd = New OleDbCommand(sql, oConnection)
comd.ExecuteNonQuery()
oConnection.Close()
else
HiddenField1.Value = "1"
.......
.......
end if
end sub 4 1114
Harry,
From what you have written, it *sounds* like what you want to do is a
cross-page postback. Look up the term "cross-page postback" and you will find
examples and description of how to use.
Peter
--
Site: http://www.eggheadcafe.com
UnBlog: http://petesbloggerama.blogspot.com
Short urls & more: http://ittyurl.net
"Harry" wrote:
Hi,
This aspx page (let's call it thispage.aspx) fetches data from a
sqldatasource, then performs several things (in code-behind) and, to
simplify, passes data from code-behind via a hiddenfield to a javascript in
the aspx file. This javascript performs things and finally send data via a
form to another database.
My problem is that when the page is postback (with this java-line:
document.getElementById("ins").action="thispage.as px"), instead of
performing the code after "If Page.IsPostBack Then", it shows the original
aspx file again.
Any idea what i have to change to do what i want to do?
Thanks
Harry
aspx file:
-------
<form id="form1" runat="server">
<asp:SqlDataSource ID="SqlDataSource1" runat="server"
ConnectionString="Provider = ... ; Data Source =...;">
</asp:SqlDataSource>
<asp:HiddenField ID="HiddenField1" runat="server" />
</form>
<form id="ins" style="width:750px;background-color:Gray" method="post">
<input id="sql" name="sql" type="hidden" />
<input id="conn" name="conn" type="hidden" />
<input runat="server" id="Submit1" type="button" value="Klik hier om
te bewaren" onclick="sendtodb()"/>
</form>
<script language="javascript" type="text/javascript">
function sendtodb()
{
var nfieldout=document.getElementById("hiddenfield1"). value
......
......
document.getElementById("sql").value=inscomm
document.getElementById("conn").value=conn
document.getElementById("ins").action="thispage.as px"
document.getElementById("ins").submit()
return true;
</script>
code-behind
-----------
Protected Sub Page_Load(ByVal sender As Object, ByVal e As
System.EventArgs) Handles Me.Load
If Page.IsPostBack Then
Dim conn, sql
sql = Request.Form("sql")
conn = Request.Form("conn")
Dim oConnection As OleDbConnection
Dim sConnection As String
oConnection = New OleDbConnection()
Dim comd As OleDbCommand
sConnection = conn
oConnection.ConnectionString = sConnection
oConnection.Open()
comd = New OleDbCommand(sql, oConnection)
comd.ExecuteNonQuery()
oConnection.Close()
else
HiddenField1.Value = "1"
.......
.......
end if
end sub
Peter, thanks for replying.
Maybe my explanation was not good, but i don't think it's cross-page
postback, since the 'action' method in javascript of the form
(document.getElementByI("ins").action="thispage.as px") refers to itself
("thispage.aspx"). The whole code here below is contained in "thispage.aspx"
and "thispage.aspx.vb".
"Peter Bromberg [C# MVP]" <pb*******@yahoo.yabbadabbadoo.comschreef in
bericht news:60**********************************@microsof t.com...
Harry,
From what you have written, it *sounds* like what you want to do is a
cross-page postback. Look up the term "cross-page postback" and you will
find
examples and description of how to use.
Peter
--
Site: http://www.eggheadcafe.com
UnBlog: http://petesbloggerama.blogspot.com
Short urls & more: http://ittyurl.net
"Harry" wrote:
>Hi,
This aspx page (let's call it thispage.aspx) fetches data from a sqldatasource, then performs several things (in code-behind) and, to simplify, passes data from code-behind via a hiddenfield to a javascript in the aspx file. This javascript performs things and finally send data via a form to another database.
My problem is that when the page is postback (with this java-line: document.getElementById("ins").action="thispage.a spx"), instead of performing the code after "If Page.IsPostBack Then", it shows the original aspx file again.
Any idea what i have to change to do what i want to do? Thanks Harry
aspx file: ------- <form id="form1" runat="server"> <asp:SqlDataSource ID="SqlDataSource1" runat="server" ConnectionString="Provider = ... ; Data Source =...;"> </asp:SqlDataSource> <asp:HiddenField ID="HiddenField1" runat="server" /> </form>
<form id="ins" style="width:750px;background-color:Gray" method="post"> <input id="sql" name="sql" type="hidden" /> <input id="conn" name="conn" type="hidden" /> <input runat="server" id="Submit1" type="button" value="Klik hier om te bewaren" onclick="sendtodb()"/> </form>
<script language="javascript" type="text/javascript">
function sendtodb() { var nfieldout=document.getElementById("hiddenfield1"). value ...... ...... document.getElementById("sql").value=inscomm document.getElementById("conn").value=conn document.getElementById("ins").action="thispage.a spx" document.getElementById("ins").submit() return true; </script>
code-behind ----------- Protected Sub Page_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load If Page.IsPostBack Then Dim conn, sql sql = Request.Form("sql") conn = Request.Form("conn") Dim oConnection As OleDbConnection Dim sConnection As String oConnection = New OleDbConnection() Dim comd As OleDbCommand sConnection = conn oConnection.ConnectionString = sConnection oConnection.Open() comd = New OleDbCommand(sql, oConnection) comd.ExecuteNonQuery() oConnection.Close() else HiddenField1.Value = "1" ....... ....... end if end sub
IsPostBack just checks for the "__Viewstate" hidden field in the
postback form fields. as you are submitting a different form on the page
then the one containing the viewstate, the viewstate hidden field is not
included in the postback data
-- bruce (sqlwork.com)
Harry wrote:
Peter, thanks for replying.
Maybe my explanation was not good, but i don't think it's cross-page
postback, since the 'action' method in javascript of the form
(document.getElementByI("ins").action="thispage.as px") refers to itself
("thispage.aspx"). The whole code here below is contained in "thispage.aspx"
and "thispage.aspx.vb".
"Peter Bromberg [C# MVP]" <pb*******@yahoo.yabbadabbadoo.comschreef in
bericht news:60**********************************@microsof t.com...
>Harry, From what you have written, it *sounds* like what you want to do is a cross-page postback. Look up the term "cross-page postback" and you will find examples and description of how to use. Peter
-- Site: http://www.eggheadcafe.com UnBlog: http://petesbloggerama.blogspot.com Short urls & more: http://ittyurl.net
"Harry" wrote:
>>Hi,
This aspx page (let's call it thispage.aspx) fetches data from a sqldatasource, then performs several things (in code-behind) and, to simplify, passes data from code-behind via a hiddenfield to a javascript in the aspx file. This javascript performs things and finally send data via a form to another database.
My problem is that when the page is postback (with this java-line: document.getElementById("ins").action="thispage. aspx"), instead of performing the code after "If Page.IsPostBack Then", it shows the original aspx file again.
Any idea what i have to change to do what i want to do? Thanks Harry
aspx file: ------- <form id="form1" runat="server"> <asp:SqlDataSource ID="SqlDataSource1" runat="server" ConnectionString="Provider = ... ; Data Source =...;"> </asp:SqlDataSource> <asp:HiddenField ID="HiddenField1" runat="server" /> </form>
<form id="ins" style="width:750px;background-color:Gray" method="post"> <input id="sql" name="sql" type="hidden" /> <input id="conn" name="conn" type="hidden" /> <input runat="server" id="Submit1" type="button" value="Klik hier om te bewaren" onclick="sendtodb()"/> </form>
<script language="javascript" type="text/javascript">
function sendtodb() { var nfieldout=document.getElementById("hiddenfield1"). value ...... ...... document.getElementById("sql").value=inscomm document.getElementById("conn").value=conn document.getElementById("ins").action="thispage. aspx" document.getElementById("ins").submit() return true; </script>
code-behind ----------- Protected Sub Page_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load If Page.IsPostBack Then Dim conn, sql sql = Request.Form("sql") conn = Request.Form("conn") Dim oConnection As OleDbConnection Dim sConnection As String oConnection = New OleDbConnection() Dim comd As OleDbCommand sConnection = conn oConnection.ConnectionString = sConnection oConnection.Open() comd = New OleDbCommand(sql, oConnection) comd.ExecuteNonQuery() oConnection.Close() else HiddenField1.Value = "1" ....... ....... end if end sub
Thanks for the explanation.
Now, if you would have a solution in mind for that, it would be great.
"bruce barker" <no****@nospam.comschreef in bericht
news:%2****************@TK2MSFTNGP04.phx.gbl...
IsPostBack just checks for the "__Viewstate" hidden field in the postback
form fields. as you are submitting a different form on the page then the
one containing the viewstate, the viewstate hidden field is not included
in the postback data
-- bruce (sqlwork.com)
Harry wrote:
>Peter, thanks for replying.
Maybe my explanation was not good, but i don't think it's cross-page postback, since the 'action' method in javascript of the form (document.getElementByI("ins").action="thispage.a spx") refers to itself ("thispage.aspx"). The whole code here below is contained in "thispage.aspx" and "thispage.aspx.vb".
"Peter Bromberg [C# MVP]" <pb*******@yahoo.yabbadabbadoo.comschreef in bericht news:60**********************************@microsof t.com...
>>Harry, From what you have written, it *sounds* like what you want to do is a cross-page postback. Look up the term "cross-page postback" and you will find examples and description of how to use. Peter
-- Site: http://www.eggheadcafe.com UnBlog: http://petesbloggerama.blogspot.com Short urls & more: http://ittyurl.net
"Harry" wrote:
Hi,
This aspx page (let's call it thispage.aspx) fetches data from a sqldatasource, then performs several things (in code-behind) and, to simplify, passes data from code-behind via a hiddenfield to a javascript in the aspx file. This javascript performs things and finally send data via a form to another database.
My problem is that when the page is postback (with this java-line: document.getElementById("ins").action="thispage .aspx"), instead of performing the code after "If Page.IsPostBack Then", it shows the original aspx file again.
Any idea what i have to change to do what i want to do? Thanks Harry
aspx file: ------- <form id="form1" runat="server"> <asp:SqlDataSource ID="SqlDataSource1" runat="server" ConnectionString="Provider = ... ; Data Source =...;"> </asp:SqlDataSource> <asp:HiddenField ID="HiddenField1" runat="server" /> </form>
<form id="ins" style="width:750px;background-color:Gray" method="post"> <input id="sql" name="sql" type="hidden" /> <input id="conn" name="conn" type="hidden" /> <input runat="server" id="Submit1" type="button" value="Klik hier om te bewaren" onclick="sendtodb()"/> </form>
<script language="javascript" type="text/javascript">
function sendtodb() { var nfieldout=document.getElementById("hiddenfield1"). value ...... ...... document.getElementById("sql").value=inscomm document.getElementById("conn").value=conn document.getElementById("ins").action="thispage .aspx" document.getElementById("ins").submit() return true; </script>
code-behind ----------- Protected Sub Page_Load(ByVal sender As Object, ByVal e As System.EventArgs) Handles Me.Load If Page.IsPostBack Then Dim conn, sql sql = Request.Form("sql") conn = Request.Form("conn") Dim oConnection As OleDbConnection Dim sConnection As String oConnection = New OleDbConnection() Dim comd As OleDbCommand sConnection = conn oConnection.ConnectionString = sConnection oConnection.Open() comd = New OleDbCommand(sql, oConnection) comd.ExecuteNonQuery() oConnection.Close() else HiddenField1.Value = "1" ....... ....... end if end sub This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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