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ajax response assign to the text fields

Posted on 2008-06-17
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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
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Question by:newbie27
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3 Comments
 
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Author Comment

by:newbie27
Comment Utility

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

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Accepted Solution

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hielo earned 500 total points
Comment Utility
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  
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LVL 8

Author Closing Comment

by:newbie27
Comment Utility
thanks
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