java - 如何解析 Android XML 中的同名标签？

Question

如何解析 XML 中的相同名称标签？

这是 XML 文件的一部分:

<?xml version="1.0" encoding="utf-8"?>
<item>  
    <RecipeDescription>A much requested chicken recipe! Easy to double for a large group. Delicious</RecipeDescription>

    <RecipeIngredientsName>1 tablespoon cornstarch</RecipeIngredientsName>
    <RecipeIngredientsName>1 tablespoon cold water</RecipeIngredientsName>
    <RecipeIngredientsName>1/2 cup white sugar</RecipeIngredientsName>
</item>

我的 XML 解析器类:

private void ProcessXml(Document data) {
    if (data != null) {
        recipesItems =new ArrayList<>();
        Element root = data.getDocumentElement();
        Node channel = root.getChildNodes().item(1);
        NodeList items = channel.getChildNodes();

        for (int i = 0; i < items.getLength(); i++) {
            Node cureentchild = items.item(i);

            if (cureentchild.getNodeName().equalsIgnoreCase("item")) {
                RecipesItem item = new RecipesItem();
                NodeList itemchilds = cureentchild.getChildNodes();

                for (int j = 0; j < itemchilds.getLength(); j++) {
                    Node cureent = itemchilds.item(j);
                    if (cureent.getNodeName().equalsIgnoreCase("RecipeDescription")) {
                        item.setRecipeDescription(cureent.getTextContent());
                    }else if (cureent.getNodeName().equalsIgnoreCase("RecipeIngredientsName")){
                        item.setRecipeIngredientsName(cureent.getTextContent());
                    }

                }
                recipesItems.add(item);
            }
        }
    }
}

public Document Getdata() {
    try {
        url = new URL(address);
        HttpURLConnection connection = (HttpURLConnection) url.openConnection();
        connection.setRequestMethod("GET");
        InputStream inputStream = connection.getInputStream();
        DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
        DocumentBuilder builder = builderFactory.newDocumentBuilder();
        Document xmlDoc = builder.parse(inputStream);
        return xmlDoc;
    } catch (Exception e) {
        e.printStackTrace();
        return null;
    }
}

但问题在于该项目中的 RecipeIngredientsName 名称相同。我怎样才能得到这些值？非常感谢。

Answer 1

如果xml文件中的内容是Static，那么可以使用以下方法:

文件位置是:res/values/filename.xml

文件名是任意的。元素的名称将用作资源 ID。

示例:

保存在 res/values/strings.xml 的 XML 文件:

 <?xml version="1.0" encoding="utf-8"?>
<resources>
    <string-array name="planets_array">
        <item>Mercury</item>
        <item>Venus</item>
        <item>Earth</item>
        <item>Mars</item>
    </string-array>
</resources>

此应用程序代码检索字符串数组:

Resources res = getResources();
String[] planets = res.getStringArray(R.array.planets_array);