我已经看到了许多关于如何在ArrayCollection中删除重复项的示例,但我似乎无法将其移植到XMLList中。在大多数ArrayCollection示例中,该示例将数组的键与hasOwnProperty方法进行比较并返回一个布尔值。这很好,但是当我使用XMLList时,我会拿什么来比较呢?假设我有:
<fx:XML id="testXML" xmlns="">
<universe>
<category cname="cat 1">
<item iname = "All"/>
<item iname = "item 1"/>
<item iname = "item 2"/>
</category>
<category cname="cat 2">
<item iname = "All"/>
<item iname = "item 3"/>
<item iname = "item 4"/>
</category>
</universe>
</fx:XML>actionscript
var myList:XMLList = testXML..@iname;将给出项"ALL“的两次出现。我知道我可能必须将XMLList转换为XMLListCollection才能使用filterFunction (我该怎么做--或者我应该从一开始就将myList定义为XMLListCollection )。然后转到filterFunction:
private function remove Duplicate (item:Object): Boolean
{
here I don't know how to compare the item to tell me if the object already exist
or not. I guess I need to compare the item to a copy of the list and see if the
item has already been seen in the copy of the list. Or is there a clean way to
do this?
}然后将所有这些内容传递给一个dropDownList:
<s:DropDownList id="myDDL" dataProvider="{myList}" />发布于 2011-05-23 20:48:25
例如,使用E4x filter function,您可以将密钥放入Object (如果密钥是字符串,否则使用字典而不是Object),并查看它是否已经存在以构建您的XMLList:
var xml:XML=<universe>
<category cname="cat 1">
<item iname="All"/>
<item iname="item 1"/>
<item iname="item 2"/>
</category>
<category cname="cat 2">
<item iname="All"/>
<item iname="item 3"/>
<item iname="item 4"/>
</category>
</universe>
function filter(xml:XML):XMLList {
var seen:Object={}
return xml..@iname.(!seen[valueOf()]&&(seen[valueOf()]=true))
}
trace( filter(xml) )这里有一个在wonderfl上的活例子:http://wonderfl.net/c/10xr
发布于 2011-05-23 20:37:17
我知道的最简单的方法就是使用ActionLinq。这段代码将获取e4x代码并将其转换为Enumerable,将属性转换为字符串,使列表中的项与众不同,并将其转储为ArrayCollection。
myList = Enumerable.from(testXML..@iname)
.cast(String)
.distinct()
.toArrayCollection();如果不想使用ActionLinq,可以使用Dictionary来实现
[Bindable]
private var myList:ArrayList;
private function removeDuplicates(data:XMLList):ArrayList {
var result:ArrayList = new ArrayList();
var found:Dictionary = new Dictionary();
for each(var item:String in data) {
if(item in found) {
continue;
}
found[item] = true;
result.addItem(item);
}
return result;
}然后,当XML准备就绪时,您可以调用它:
myList = removeDuplicates(testXML..@iname);https://stackoverflow.com/questions/6096837
复制相似问题