Web Development

Drag and drop the contents of a particular column in a table.

30-Nov-06 15:34

You have to create an IIS web site that uses that port, 1484. The default web site has the defaut port of 80, http://localhost:80/.

_______________________29-Nov-06 18:05

_______________________

29-Nov-06 18:05

There are 3 rows in a table. The columns of the last row are filled with alphabets. I should be able to drag any alphabet (from the last row) and drop in in any of the first 2 rows. Undoing of this move should be possible. Can anybody help me out?

Re: Drag and drop the contents of a particular column in a table.

Private_Void1-Dec-06 11:16

Private_Void

1-Dec-06 11:16

Someone may be able to provide assistance if you give some context in which you are speaking. For example drag "alphabet" from last row and drop it in any of the first 2 rows. Where are these rows and from where are you dragging and dropping?

Mapping a network drive over the net

TheJudeDude29-Nov-06 13:03

29-Nov-06 13:03

HI all!
I hope someone can point me in the right direction. Here is my delima.
Our company wants two locations to share data. The data will be stored on our web server. The program that needs to access this data will be on the local computers and needs to pointed to a lettered drive. I have a webfolder setup in "My Network Places" but cannot map the drive. Is there a good way to work around this?
Any help will be very much appreciated!

Jude

George L. Jackson29-Nov-06 13:46

29-Nov-06 13:46

You have to create a network share, Network Sharing and Security, (not a "web share") on the server that hosts your web server. Also, you have to set up the proper security access.

TheJudeDude29-Nov-06 14:05

29-Nov-06 14:05

OK..I've done that. How do I map it?
EX: When I go to map network drive, you get a chance to input the info. I put in "http://"our IP/the web folder/ the shared folder, but get a "Can not be found" error.

Jude

Dave Kreskowiak29-Nov-06 14:10

Dave Kreskowiak

29-Nov-06 14:10

You can't use http:// to map a drive. You have to specify it in the format as \\ipaddress\shareName

I hope you're security is setup tight and the machine is patched up as current as 1 minute ago because you're doing something that is VERY dangerous over the Internet out in the open.

Depending on how sensitive the data is, you'd probably be better served by setting up a VPN between the two networks.

Dave Kreskowiak
Microsoft MVP - Visual Basic

George L. Jackson29-Nov-06 14:44

29-Nov-06 14:44

Jude,

I am sorry! I was called away before completing my answer. Dave is right! If you share data via the internet this way, you are exposing your web server to security breaches. As Dave said, VPN will serve you better.

TheJudeDude30-Nov-06 5:27

30-Nov-06 5:27

Thanx for all the input. I setup a VPN on the server, but cannot connect. Here is what I choose step by step:
Remote Access
VPN
The network card that is connected to the router-->Modem
From a selected range (192.168.1.220-222)(way outside of dhcp request)
Use routing and remote access
Then I get this message:
To support the relaying of DHCP messages from remote access clients,you must configure the proerties of the DHCP Relay Agent with the ip addressof you DHCP server.
I inputed the ip address of our dhcp server.
Then I created a VPN connection on the client, but it will not connect (Error 800)??
I know this is asking alot, and I am all but googled out. Please help me o great ones...

Jude

George L. Jackson30-Nov-06 10:57

30-Nov-06 10:57

I am a programmer with a little networking knowledge (enough to get by). You probably get a useful response from a non-programming discussion board.

TheJudeDude30-Nov-06 11:50

30-Nov-06 11:50

All help is greatly apporeciated..I am in the same boat as you...

Jude

IE7 Javascript Problem

Infomine Liam29-Nov-06 12:49

Infomine Liam

29-Nov-06 12:49

I'm trying to do some form validation that involves looping through all elements in a form. It works perfectly in FireFox, but totally chokes in IE7. It seems to be that the $Form.elements.length value is not returning a number.
I've written a little test function to simulate the error.

<br />
function validateForm()<br />
{<br />
  alert(document.forms.length);<br />
  var numElements = document.forms[0].elements.length;<br />
  alert(numElements);<br />
}<br />

First alert shows: 1
Second alert shows: [object]

Anyone seen this happen before?

Thanks a lot. Cool | :cool:

Torsten Mauz30-Nov-06 1:49

Torsten Mauz

30-Nov-06 1:49

try something like this

function checkForm()
{
    var frm = document.getElementById('form1');
    var strError = 'The following fields are required:\n';
    var bValid = true;
    var child;
    
    for(var i = 0; i < frm.childNodes.length; i++)
    {
        child = frm.childNodes.item(i);
        if(child.nodeType == 1)
        {
            if(child.tagName.toString().toLowerCase() == 'input')
            {
                if(child.type.toString().toLowerCase() == 'text')
                {
                    if(child.value == '')
                    {
                        strError += '\n'+ child.id.toString();
                        bValid = false;
                    }
                }
            }
        }
    }
    if(!bValid)
        alert(strError);
    return bValid;
}

Works in both FF and IE7. It only checks input text's at the moment so you'll need to change it so that it's a bit more useful, but it should get you going.

HTH

Infomine Liam30-Nov-06 6:11

Infomine Liam

30-Nov-06 6:11

Thanks very much for the reply, but I guess my main question here is why $Form.elements.length is not returning a number. This has always worked before....

Torsten Mauz30-Nov-06 6:36

Torsten Mauz

30-Nov-06 6:36

Wierd, just done a tiny test myself using your code and get a number as you'd expect from elements.length

Here is the code I ran:

<form id="form1" name="form1" onsubmit="return checkForm();">
<input type="text" id="Text1" />
<input type="text" id="Text2" />
<input type="text" id="Text3" />
<input type="text" id="Text4" />
<input type="submit" />
</form>
<script>
function checkForm()
{
alert(document.forms.length);
var numElements = document.forms[0].elements.length;
alert(numElements);
return true;
}
</script>

The first alert gives "1" and the second gives "5". If your form isn't too huge perhaps you could paste in the code that's giving you this wierd behaviour.

HTH

Infomine Liam30-Nov-06 6:53

Infomine Liam

30-Nov-06 6:53

Well, I feel dumb now. I had a textbox in the form named "length". D'Oh! | :doh:

Thanks for all your help.