Question

我在ASP Classic中解析并显示RSS源，我想按字母顺序按特定节点对项目进行排序。

Set xmlDOM = Server.CreateObject("MSXML2.DOMDocument")
xmlDOM.async = False

xmlDOM.setProperty "ServerHTTPRequest", True
xmlDOM.Load("http://myfeedhere.xml")

Set itemList = xmlDOM.getElementsByTagName("item")

''# Then I am getting the values of each node this way:

For Each item in itemList
    For Each child in item.childNodes
        Select case lcase(child.nodeName)
            case "title"
                title = child.text

            case "link"
                link = child.text

            case "fname"
                fname = child.text

            case "lname"
               lname = child.text

            case "media:content"
                media = child.getAttribute("url")
        End Select
    next

我需要通过lname“节点对itemList进行排序，这样做的最佳方法是什么..

当我只需要两个节点时，添加标题和链接到字典对象。我在按键数组上调用了quicksort然后相应地输出。此外，我必须为重复的姓氏做好准备，这意味着姓氏不能成为关键。

Answer 1

得到帮助并非常感谢http://www.4guysfromrolla.com/webtech/012799-3.shtml 用于二维快速排序功能我制定了一个解决方案（除了告诉客户更新到.net和C＃）

...

dim xmlDOM,itemList,currentLetter
Set xmlDOM = Server.CreateObject("MSXML2.DOMDocument")  
xmlDOM.async = False  

xmlDOM.setProperty "ServerHTTPRequest", True  
xmlDOM.Load("http://yourrssfeedhere.xml")


    Set itemList = xmlDOM.getElementsByTagName("item")

    dim itemListLength 
    itemListLength = itemList.length

    dim lastIndex
     lastIndex = itemListLength -1 
   'get the value of the last index of all the items

    dim byLastName()

    redim byLastName(lastIndex,3)

  ' set two-dimensinal array indexes, ([lastIndexofItemlist],[#of fields])


    dim it

    For Each item in itemList
    itemCount = itemCount + 1




    For Each child in item.childNodes


    Select case lcase(child.nodeName)

' loop through each node of item and set variable, set each case for your specific nodes in each <item>
' (i.e <category>,<lname>,etc)

     case "title"
       title = child.text

     case "description"
       description = child.text

     case "link"
       link = child.text

     case "pubdate"
       pubdate = child.text


     case "fname"
       fname = child.text

     case "lname"
      lname = child.text

      case "media:content"
      media = child.getAttribute("url")


     End Select



     next

     i = itemCount-1 



     byLastName(i,0) = title
     byLastName(i,1) = link
     byLastName(i,2) = lname
     byLastName(i,3) = fname


    next



    dim namesLength
    namesLength = Ubound(byLastName) 

 'last index of bylastname

for n = 0 to namesLength
' simple output before sort
Response.Write("<p> Last Name :" & byLastName(n,2) & ", First Name :" & byLastName(n,3) &"</p>")
next

Response.Write("<p> Sorted:</p>")



 Call QuickSort(byLastName,0,ubound(byLastName,1),2) 

'the last parameter is the field to sort the array on ,
' in this case 2 = lname
    for n = 0 to namesLength
    ' output to test sort
    Response.Write("<p> Last Name :" & byLastName(n,2) & ", First Name :" & byLastName(n,3) &"</p>")

    next

Parsin - 在ASP Classic中对多维数组或对象列表中的值或属性进行排序

1 个答案: