Solved

ajax response assign to the text fields

Posted on 2008-06-17
3
222 Views
Last Modified: 2010-04-21
Hello Experts,
I am using these function to do ajax request to the builder.asp which originally has returned comma separated list ids,
I now want to return 2 more information ie: name and notes from the showListRefs function and assign it to the corresponding text fields.
Please can you help
Thanks
Sam
0
Comment
Question by:newbie27
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 2
3 Comments
 
LVL 8

Author Comment

by:newbie27
ID: 21805399

javascript function
----------------------
function getListRefs(sListname)
{
	$.ajax({
	  type: "POST",
	  url: "list_builder.asp",
	  data: "action=results&listname="+sListname,
	  success: function(msg)
	  		{					
				$("#listRefnos").val( msg );		
	
				var sRefNumbers = $("#listRefnos").val();
				var sViewQuery = "SF1=keyword&ST1="+sRefNumbers;     
				//alert("sViewQuery: "+sViewQuery);
				
				listViewer("#listDisplay1",sViewQuery);
				
                                $("#txtListName").val('test6');     //  I want to assign the name here
				$("#txtNotes").val('test notes');   //  and notes here
			}
	 
	});
} 
 
 
ASP function
-------------
 
 
 
PRIVATE FUNCTION showListRefs(sListName)
    sFilename = "test@test.com.xml"      	
    sXMLFile = ADMIN_WWW_FOLDER & "\lists\data\" & sFilename            
    Set objXML = Server.CreateObject("Microsoft.XMLDOM")
 
	If objXML.load(sXMLFile) Then
		Set objRoot = objXML.documentElement
    End if
 
    Set oNodes = objXML.selectNodes("//lists/*")
   
	count = 1
	For Each oChild In oNodes
	 
		if oChild.childNodes(0).text = sListName Then
		   	showListRefs = oChild.childNodes(2).text    '   this will give us comma separeted IDs
                       
                        'showListRefs =   oChild.childNodes(1).text   ' this will give the list name	 
                        'showListRefs =   oChild.childNodes(0).text   ' this will give the list notes
			
		End if
	Next 
	showListRefs = showListRefs 
 
END Function

Open in new window

0
 
LVL 82

Accepted Solution

by:
hielo earned 500 total points
ID: 21811930
You could pipe separate the data for each record. Intead of just:
showListRefs = oChild.childNodes(2).text    

try:
showListRefs = oChild.childNodes(2).text & "|" & oChild.childNodes(1).text   & "|" & oChild.childNodes(0).text  
0
 
LVL 8

Author Closing Comment

by:newbie27
ID: 31468058
thanks
0

Featured Post

Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

This article shows how to create and access 2-dimensional arrays in JavaScript.  It includes a tutorial in case you are just trying to "get your head wrapped around" the concept and we'll also look at some useful tips for more advanced programmers. …
In this article, we'll look how to sort an Array in JavaScript, including the more advanced techniques of sorting a collection of records either ascending or descending on two or more fields. Basic Sorting of Arrays First, let's look at the …
The viewer will learn the basics of jQuery, including how to invoke it on a web page. Reference your jQuery libraries: (CODE) Include your new external js/jQuery file: (CODE) Write your first lines of code to setup your site for jQuery.: (CODE)
The viewer will learn the basics of jQuery including how to code hide show and toggles. Reference your jQuery libraries: (CODE) Include your new external js/jQuery file: (CODE) Write your first lines of code to setup your site for jQuery…
Suggested Courses

739 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question