Question

我对XML和Android开发有点新...我遇到过这个问题，我需要解析一个XML，其中的元素是相同的，并包含整个元素。这有点难以解释，请参阅下面的代码：

<tns:camera>

<tns:congestionLocations>
<tns:congestion>Free Flow</tns:congestion>
<tns:direction>Eastbound</tns:direction>
</tns:congestionLocations>

<tns:congestionLocations>
<tns:congestion>Free Flow</tns:congestion>
<tns:direction>Westbound</tns:direction>
</tns:congestionLocations>


<tns:description>Bond St looking east</tns:description>
<tns:direction>Eastbound</tns:direction>
<tns:group>SH16-North-Western</tns:group>
<tns:lat>-36.869</tns:lat>
<tns:lon>174.746</tns:lon>
<tns:name>SH16 1 Bond St</tns:name>
<tns:viewUrl>http://www.trafficnz.info/camera/view/130</tns:viewUrl>
</tns:camera>

基本上，我需要解析整个元素（tns：camera）并包含拥塞位置（显然彼此分开），但是在同一个类中，我将在列表视图中使用所有这些... < / p>

我将如何实现这一目标？

目前，我正在使用Pull Parser，并将其解析为类对象

PullParser代码：

case XmlPullParser.END_TAG:
                    if (tagname.equalsIgnoreCase(KEY_SITE)) {current Site
                        CameraSites.add(curCameraClass);
                    } else if (tagname.equalsIgnoreCase(KEY_DESCRIPTION)) {

                        curCameraClass.setDescription(curText);
                    }else if (tagname.equalsIgnoreCase(KEY_NAME)) {
                        curCameraClass.setName(curText);
                    }

                    break;

亲切的问候！

Answer 1

你可以SAXParser来解析xml。希望以下链接有所帮助：

developersite

basic tutorial

Answer 2

试试这个..

        NodeList nodeList = doc.getElementsByTagName("tns:camera");

                for (int i = 0; i < nodeList.getLength(); i++) {

                    Node node = nodeList.item(i);       

                    Element fstElmnt = (Element) node;
                    NodeList nameList = fstElmnt.getElementsByTagName("tns:group");
                    Element nameElement = (Element) nameList.item(0);
                    nameList = nameElement.getChildNodes();

                    System.out.println("tns:group : "+((Node) nameList.item(0)).getNodeValue());


                    Element fstElmnt1 = (Element) node;
                    NodeList nameList1 = fstElmnt1.getElementsByTagName("tns:viewUrl");
                    Element nameElement1 = (Element) nameList1.item(0);
                    nameList1 = nameElement1.getChildNodes();

                    System.out.println("tns:viewUrl : "+ ((Node) nameList1.item(0)).getNodeValue());

//same as use to all tns:description,tns:direction and tns:lat etc.,


                    if(node.getNodeType() == Node.ELEMENT_NODE)
                    {
                        Element e = (Element) node;
                        NodeList resultNodeList = e.getElementsByTagName("tns:congestionLocations");
                        int resultNodeListSize = resultNodeList.getLength();
                        for(int j = 0 ; j < resultNodeListSize ; j++ )
                        {
                            Node resultNode = resultNodeList.item(j);
                            if(resultNode.getNodeType() == Node.ELEMENT_NODE)
                            {
                                Element fstElmnt2 = (Element) resultNode;
                                NodeList nameList2 = fstElmnt2.getElementsByTagName("tns:congestion");
                                Element nameElement2 = (Element) nameList2.item(0);
                                nameList2 = nameElement2.getChildNodes();

                                Log.v("tns:congestion", ""+((Node) nameList2.item(0)).getNodeValue());

                                Element fstElmnt3 = (Element) resultNode;
                                NodeList nameList3 = fstElmnt3.getElementsByTagName("tns:direction");
                                Element nameElement3 = (Element) nameList3.item(0);
                                nameList3 = nameElement3.getChildNodes();

                                Log.v("tns:direction--", ""+((Node) nameList3.item(0)).getNodeValue());
                            }
                        }
                    }

                }

解析具有相同名称android的两个元素

2 个答案: