我有一个HTML文件,例如:http://scholar.google.gr/citations?user=v9xULZwAAAAJ&hl=el
在此文件中,存在带有文章的表。我想使用xpath获得前20篇文章(如果存在)。
我尝试找到拳头文章:
String str = (String) xpath.evaluate("//form[contains(@id,'citationsForm')]/div[2]/div[1]/table/tbody/tr[2]/td[@id='col-title']/a", docList.get(0), XPathConstants.STRING);
好吧!结果:现代信息检索
对于所有文章:
String str = (String) xpath.evaluate("//form[contains(@id,'citationsForm')]/div[2]/div[1]/table/tbody/tr/td[@id='col-title']/a", docList.get(0), XPathConstants.STRING);
但不起作用
任何的想法?
比你!
编辑:
我也尝试:
NodeList result = (NodeList)xpath.evaluate("//form[contains(@id,'citationsForm')]/div[2]/div[1]/table/tbody/tr/td[@id='col-title']/a",
docList.get(0), XPathConstants.NODESET);
ArrayList<String>liste = new ArrayList<String>();
for(int i=0; i<result.getLength();i++){
System.out.println(result.item(i).getNodeValue());
liste.add(result.item(i).getNodeName());
}
编辑2所有代码
类FileOperation:
package xmlparse;
import java.io.BufferedReader;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.logging.Level;
import java.util.logging.Logger;
import javax.xml.parsers.ParserConfigurationException;
import org.htmlcleaner.CleanerProperties;
import org.htmlcleaner.DomSerializer;
import org.htmlcleaner.HtmlCleaner;
import org.htmlcleaner.TagNode;
import org.w3c.dom.Document;
public class FileOperations {
private static final String path = "C:\\Users\\Dimitris\\Desktop\\authors";
public ArrayList<Document> getXmlDocumt() {
ArrayList<Document> xmlFileList = new ArrayList<>();
try {
ArrayList<File> listFiles = listFiles(path);
for (File f : listFiles) {
String html = readfile(f.getAbsolutePath());
xmlFileList.add(ConvertHtml2Xml(html) );
}
} catch (IOException ex) {
Logger.getLogger(FileOperations.class.getName()).log(Level.SEVERE, null, ex);
}
return xmlFileList;
}
private ArrayList<File> listFiles(String directoryName) throws IOException {
ArrayList<File> htmlfilelist = new ArrayList<>();
File directory = new File(directoryName);
//get all the files from a directory
File[] fList = directory.listFiles();
for (File file : fList) {
if (file.isFile()) {
htmlfilelist.add(file);
}
}
return htmlfilelist;
}
private String readfile(String file) throws FileNotFoundException, IOException {
String s = "";
FileReader fr = new FileReader(file);
BufferedReader br = new BufferedReader(fr);
StringBuilder content = new StringBuilder(1024);
while ((s = br.readLine()) != null) {
content.append(s);
}
//System.out.println(content.toString());
return content.toString();
}
private Document ConvertHtml2Xml(String html) {
TagNode tagNode = new HtmlCleaner().clean(html);
Document doc = null;
try {
doc = new DomSerializer(new CleanerProperties()).createDOM(tagNode);
} catch (ParserConfigurationException ex) {
Logger.getLogger(FileOperations.class.getName()).log(Level.SEVERE, null, ex);
}
return doc;
}
}
XpathQueries类:
XPath xpath;
ArrayList<Document> docList;
public XpathQueries() {
xpath = XPathFactory.newInstance().newXPath();
FileOperations fo = new FileOperations();
docList = new ArrayList<>(fo.getXmlDocumt());
}
public void getArticle() throws XPathExpressionException {
// String str = (String) xpath.evaluate("//form[contains(@id,'citationsForm')]/div[2]/div[1]/table/tbody//td[1]/a",
// docList.get(0), XPathConstants.STRING);
String str = (String) xpath.evaluate("//*[@id='col-title']/a", docList.get(0), XPathConstants.STRING);
System.out.println(str);
}
}
最佳答案
试试这个:
Object result = xpath.evaluate("//*[@id='col-title']/a", docList.get(0), XPathConstants.STRING);
NodeList nodes = (NodeList) result;
for (int i = 0; i < nodes.getLength(); i++) {
System.out.println(nodes.item(i).getNodeValue());
}
关于java - java-xpath获取表内的行,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/22571425/