[英]Flatten XML document in Java
Is there any way to Flatten an XML document by removing all the nested tags / arrays to simple Key / Value pair. 有什么方法可以通过将所有嵌套的标签/数组移除为简单的键/值对来展平XML文档。 For example I have an example
例如我有一个例子
<root>
<a>
<b>some-value1</b>
<c>some-value2</c>
</a>
<a>
<b>some-value3</b>
<c>some-value4</c>
</a>
<p>some-value-p</p>
TO 至
root.a.0.b=some-value1
root.a.0.c=some-value2
root.a.1.b=some-value3
root.a.1.c=some-value4
root.p=some-value-p
I am looking for some solution in Java or Scala? 我在寻找Java或Scala的解决方案?
You could use Jackson 你可以用杰克逊
Instead of 代替
root.a.0.b=some-value1
root.a.0.c=some-value2
root.a.1.b=some-value3
root.a.1.c=some-value4
root.p=some-value-p
You will get 你会得到
#converted XML
#Tue Sep 19 15:15:57 CEST 2017
a.2.c=some-value4
a.2.b=some-value3
p=some-value-p
a.1.c=some-value2
a.1.b=some-value1
Here is test code: 这是测试代码:
import com.fasterxml.jackson.dataformat.javaprop.JavaPropsMapper;
import com.fasterxml.jackson.dataformat.xml.XmlMapper;
import com.fasterxml.jackson.dataformat.xml.annotation.JacksonXmlElementWrapper;
import java.io.IOException;
import java.io.StringReader;
import java.util.List;
import java.util.Properties;
import org.junit.Test;
public class TestXMLConversion {
public static final String XML =
"<root>\n" +
"<a>\n" +
" <b>some-value1</b>\n" +
" <c>some-value2</c>\n" +
"</a>\n" +
"<a>\n" +
" <b>some-value3</b>\n" +
" <c>some-value4</c>\n" +
"</a>\n" +
"<p>some-value-p</p>" +
"</root>";
public static class Root {
@JacksonXmlElementWrapper(useWrapping=false)
private List<A> a;
private String p;
public List<A> getA() {
return a;
}
public void setA(List<A> a) {
this.a = a;
}
public String getP() {
return p;
}
public void setP(String p) {
this.p = p;
}
}
public static class A {
private String b;
private String c;
public String getB() {
return b;
}
public void setB(String b) {
this.b = b;
}
public String getC() {
return c;
}
public void setC(String c) {
this.c = c;
}
}
@Test
public void test() throws IOException {
XmlMapper xmlMapper = new XmlMapper();
Root parsedData = xmlMapper.readValue(new StringReader(XML), Root.class);
JavaPropsMapper propertiesMapper = new JavaPropsMapper();
Properties properties = propertiesMapper.writeValueAsProperties(parsedData);
properties.store(System.out, "converted XML");
}
}
The dependencies I used: 我使用的依赖项:
<dependency>
<groupId>com.fasterxml.jackson.dataformat</groupId>
<artifactId>jackson-dataformat-xml</artifactId>
<version>2.9.1</version>
</dependency>
<dependency>
<groupId>com.fasterxml.jackson.dataformat</groupId>
<artifactId>jackson-dataformat-properties</artifactId>
<version>2.9.1</version>
</dependency>
<dependency>
<groupId>junit</groupId>
<artifactId>junit</artifactId>
<version>4.12</version>
<scope>test</scope>
</dependency>
Here is a solution without external dependencies. 这是没有外部依赖性的解决方案。 You can try it !
你可以试试看 !
import org.w3c.dom.Document;
import org.w3c.dom.Element;
import org.w3c.dom.Node;
import org.w3c.dom.NodeList;
import org.xml.sax.SAXException;
import javax.xml.parsers.DocumentBuilderFactory;
import javax.xml.parsers.ParserConfigurationException;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.nio.charset.StandardCharsets;
public class Main {
private static final String xml =
"<root>\n" +
" <a>\n" +
" <b>some-value1</b>\n" +
" <c>some-value2</c>\n" +
" </a>\n" +
" <a>\n" +
" <b>some-value3</b>\n" +
" <c>some-value4</c>\n" +
" </a>\n" +
" <p>some-value-p</p>\n" +
"</root>";
public static void main(String[] args) {
try {
byte[] bytes = xml.getBytes(StandardCharsets.UTF_8);
try (InputStream input = new ByteArrayInputStream(bytes)) {
Document document = DocumentBuilderFactory.newInstance().newDocumentBuilder().parse(input);
Element root = document.getDocumentElement();
flattXml("", root);
}
} catch (IOException | ParserConfigurationException | SAXException e) {
e.printStackTrace();
}
}
private static void flattXml(String currentPath, Node currentNode) {
if (currentNode.getNodeType() == Node.TEXT_NODE &&
!currentNode.getNodeValue().trim().isEmpty()) {
System.out.println(currentPath + "=" + currentNode.getNodeValue());
} else {
NodeList childNodes = currentNode.getChildNodes();
int length = childNodes.getLength();
String nextPath = currentPath.isEmpty()
? currentNode.getNodeName()
: currentPath + "." + currentNode.getNodeName();
for (int i = 0; i < length; i++) {
Node item = childNodes.item(i);
flattXml(nextPath, item);
}
}
}
}
Console output: 控制台输出:
root.a.b=some-value1
root.a.c=some-value2
root.a.b=some-value3
root.a.c=some-value4
root.p=some-value-p
Parse the XML, then recursively step through the node tree keeping track of the parent. 解析XML,然后递归地浏览节点树并跟踪父节点。 Then print parent + "."
然后打印父+“。 + node + "=" + value.
+节点+“ =” +值。
If you are going through the trouble of converting it though, it may be easier just to parse it directly into a useful object though using something like Jackson. 如果您遇到转换麻烦,可以使用Jackson之类的方法将其直接解析为有用的对象会更容易。 Then you can account for things like attributes in the XML.
然后,您可以考虑诸如XML中的属性之类的事情。
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