如何进入必需的标签。XML
如何进入必需的标签。XML
提问于 2019-05-23 05:24:23
我有以下xml文件：
<rss xmlns:mo="http://news.yandex.ru" version="2.0">
<channel>
<title>..</title>
<link>.</link>
<description>..</description>
<lastBuildDate>..</lastBuildDate>
<generator>..</generator>
<image>
<url>..</url>
<title>..</title>
<link>..</link>
</image>
<item>
<title></title>
<link></link>
<description></description>
<mo:full-text></mo:full-text>
<author></author>
<pubDate></pubDate>
</item>
<item>
....
</item>
Another item...till the end
</channel>
</rss>我得去拿一份物品清单。
List<Item> items = new ArrayList<>();类项目具有下一个字段：
public static class Item{
public final String titleNews;
public final String link;
public final String description;
public final String date;
public final String author;
private Item(String titleNews, String link, String description, String date, String author) {
this.titleNews = titleNews;
this.link = link;
this.description = description;
this.date = date;
this.author = author;
}
}我用https://developer.android.com/training/basics/network-ops/xml写的代码
问题是我不知道如何进入频道的标签，然后跳过无用的标签，直到项目的标签。然后将每一项添加到我的列表中。
private List<Item> readFeed(XmlPullParser parser) throws XmlPullParserException, IOException {
List<Item> items = new ArrayList<>();
parser.require(XmlPullParser.START_TAG, ns, "rss");
while (parser.next() != XmlPullParser.END_TAG) {
if (parser.getEventType() != XmlPullParser.START_TAG) {
continue;
}
String name = parser.getName();
// Starts by looking for the entry tag
if (name.equals("item")) {
items.add(readItem(parser));
} else {
skip(parser);
}
}
return items;
}回答 2
Stack Overflow用户
回答已采纳
发布于 2019-05-23 05:36:52
您已经在这样做了，但是您需要一个类似的方法来处理channel内容。
当然，您可能应该有一个Channel对象来收集所有相关的频道信息，而不仅仅是Item对象的组合列表。
private List<Item> readRss(XmlPullParser parser) throws XmlPullParserException, IOException {
List<Item> items = new ArrayList<>();
parser.require(XmlPullParser.START_TAG, ns, "rss");
while (parser.next() != XmlPullParser.END_TAG) {
if (parser.getEventType() != XmlPullParser.START_TAG) {
continue;
}
String name = parser.getName();
if (name.equals("channel")) {
items.addAll(readChannel(parser));
} else {
skip(parser);
}
}
return items;
}private List<Item> readChannel(XmlPullParser parser) throws XmlPullParserException, IOException {
List<Item> items = new ArrayList<>();
parser.require(XmlPullParser.START_TAG, ns, "channel");
while (parser.next() != XmlPullParser.END_TAG) {
if (parser.getEventType() != XmlPullParser.START_TAG) {
continue;
}
String name = parser.getName();
if (name.equals("item")) {
items.add(readItem(parser));
} else {
skip(parser);
}
}
return items;
}private Item readItem(XmlPullParser parser) throws XmlPullParserException, IOException {
// code here
}Stack Overflow用户
发布于 2019-05-23 05:38:17
我会使用xpath。https://www.baeldung.com/java-xpath
<?xml version="1.0"?>
<Tutorials>
<Tutorial tutId="01" type="java">
<title>Guava</title>
<description>Introduction to Guava</description>
<date>04/04/2016</date>
<author>GuavaAuthor</author>
</Tutorial>
<Tutorial tutId="02" type="java">
<title>XML</title>
<description>Introduction to XPath</description>
<date>04/05/2016</date>
<author>XMLAuthor</author>
</Tutorial>
</Tutorials>^ XML示例文件
vvvv Xpath代码。
FileInputStream fileIS = new FileInputStream(this.getFile());
DocumentBuilderFactory builderFactory = DocumentBuilderFactory.newInstance();
DocumentBuilder builder = builderFactory.newDocumentBuilder();
Document xmlDocument = builder.parse(fileIS);
XPath xPath = XPathFactory.newInstance().newXPath();
String expression = "/Tutorials/Tutorial";
nodeList = (NodeList) xPath.compile(expression).evaluate(xmlDocument, XPathConstants.NODESET);页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/56265217
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