整数名访问XML节点 [英] Access XML nodes with integer names

问题描述

有关我的申请，我做的HTT prequest，并取回一些XML服务从一个JSP。该XML具有一定的（是的，我知道这是无效/不正确的XML如果我不能找到一个绷带，我会尽量解决在内部。）与整数节点的名称，说＆LT; 2 - ; 为例。

当我试图访问它，使用 myXMLVariable.child（2），则返回第三个（指数= 2）XML节点代替。据我所知，这种行为是正确的 。有什么办法可以解决这个问题？

---

示例

  VAR与myXML：字符串=＆LT;响应＆gt;中+
                    ＆其中; PLACE1＆gt;中+
                    ＆LT;项目＆GT; 1＆LT; /项目＆gt;中+
                    ＆其中;东西→1＆其中; /东西＆gt;中+
                    ＆所述; / PLACE1＆gt;中+
                    2＆gt;中+
                    ＆LT;项目＆GT; 1＆LT; /项目＆gt;中+
                    ＆其中;东西→1＆其中; /东西＆gt;中+
                    ＆所述; / 2＆gt;中+
                    ＆其中; place3＆gt;中+
                    ＆LT;项目＆GT; 1＆LT; /项目＆gt;中+
                    ＆其中;东西→1＆其中; /东西＆gt;中+
                    ＆所述; / place3＆gt;中+
                    ＆所述; /响应＆gt;中;

保护功能getParam（）：无效
{
    VAR的xml：XML =新的XML（与myXML）;

    Alert.show（xml.child（2））;
    //trace(xml.child("2））
}
 

xml.child（2）返回

 ＆LT; place3＆GT;
    ...
＆LT; / place3＆GT;
 

...当我想

  2＆GT;
    ...
＆所述; / 2  - ;
 

---

注意

我知道这是无效的XML。我要寻找一个解决办法，短期修复。有一个近未来的发布日期，而这种解决办法将被删除，并用正确的XML的下一个版本取代。

解决方案

使用E4X搜索前就返回XMLList pression。

 跟踪（xml.children（）（名（）==2）toXMLString（））;
 

1. 在获取所有的孩子
2. 搜索您需要的名称（）。

For my application, I make an HTTPRequest, and get back some XML served from a JSP. That XML has some (yes, I'm aware this is invalid/improper XML. If I can't find a bandaid, I will try to address that internally) nodes with integers as names, say <2> for example.

When I attempt to access it, using myXMLVariable.child("2"), it returns the third (index=2) XML node instead. I understand that this behavior is "correct". Is there any way to get around this behavior?

---

Example

var myXML:String = "<response>" +
                    "<place1>" +
                    "   <item>1</item>" +
                    "   <stuff>1</stuff>" +
                    "</place1>" +
                    "<2>" +
                    "   <item>1</item>" +
                    "   <stuff>1</stuff>" +
                    "</2>" +
                    "<place3>" +
                    "   <item>1</item>" +
                    "   <stuff>1</stuff>" +
                    "</place3>" +
                    "</response>";

protected function getParam():void
{
    var xml:XML = new XML(myXML);

    Alert.show(xml.child("2"));
    //trace(xml.child("2"))
}

xml.child("2") returns

<place3>
    ...
</place3>

...when I want

<2>
    ...
</2>

---

NOTE

I am aware this is invalid XML. I am looking for a workaround, a short term fix. There is a near-future release date, and this workaround will be removed and replaced with proper XML for the next version.

解决方案

Use E4X search expression on XMLList.

trace(xml.children().(name() == "2").toXMLString());

1. Get all children
2. Search for the name() you need.

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