84

我正在尝试创建一个简单的 java 程序，它从 zip 文件中的文件中读取和提取内容。Zip 文件包含 3 个文件（txt、pdf、docx）。我需要阅读所有这些文件的内容，为此我正在使用Apache Tika 。

有人可以在这里帮助我实现该功能。到目前为止我已经尝试过了，但没有成功

代码片段

public class SampleZipExtract {


    public static void main(String[] args) {

        List<String> tempString = new ArrayList<String>();
        StringBuffer sbf = new StringBuffer();

        File file = new File("C:\\Users\\xxx\\Desktop\\abc.zip");
        InputStream input;
        try {

          input = new FileInputStream(file);
          ZipInputStream zip = new ZipInputStream(input);
          ZipEntry entry = zip.getNextEntry();

          BodyContentHandler textHandler = new BodyContentHandler();
          Metadata metadata = new Metadata();

          Parser parser = new AutoDetectParser();

          while (entry!= null){

                if(entry.getName().endsWith(".txt") || 
                           entry.getName().endsWith(".pdf")||
                           entry.getName().endsWith(".docx")){
              System.out.println("entry=" + entry.getName() + " " + entry.getSize());
                     parser.parse(input, textHandler, metadata, new ParseContext());
                     tempString.add(textHandler.toString());
                }
           }
           zip.close();
           input.close();

           for (String text : tempString) {
           System.out.println("Apache Tika - Converted input string : " + text);
           sbf.append(text);
           System.out.println("Final text from all the three files " + sbf.toString());
        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (IOException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (SAXException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        } catch (TikaException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
    }
}
4

6 回答 6

205

如果您想知道如何从每个文件中获取文件内容，ZipEntry这实际上非常简单。这是一个示例代码：

public static void main(String[] args) throws IOException {
    ZipFile zipFile = new ZipFile("C:/test.zip");

    Enumeration<? extends ZipEntry> entries = zipFile.entries();

    while(entries.hasMoreElements()){
        ZipEntry entry = entries.nextElement();
        InputStream stream = zipFile.getInputStream(entry);
    }
}

拥有 InputStream 后，您可以随心所欲地阅读它。

于 2013-03-27T19:05:25.660 回答
52

从 Java 7 开始，NIO Api 提供了一种更好、更通用的方式来访问 Zip 或 Jar 文件的内容。实际上，它现在是一个统一的 API，允许您将 Zip 文件与普通文件一样对待。

为了提取此 API 中 zip 文件中包含的所有文件，您可以执行以下操作：

在 Java 8 中：

private void extractAll(URI fromZip, Path toDirectory) throws IOException{
    FileSystems.newFileSystem(fromZip, Collections.emptyMap())
            .getRootDirectories()
            .forEach(root -> {
                // in a full implementation, you'd have to
                // handle directories 
                Files.walk(root).forEach(path -> Files.copy(path, toDirectory));
            });
}

在 Java 7 中：

private void extractAll(URI fromZip, Path toDirectory) throws IOException{
    FileSystem zipFs = FileSystems.newFileSystem(fromZip, Collections.emptyMap());

    for(Path root : zipFs.getRootDirectories()) {
        Files.walkFileTree(root, new SimpleFileVisitor<Path>() {
            @Override
            public FileVisitResult visitFile(Path file, BasicFileAttributes attrs) 
                    throws IOException {
                // You can do anything you want with the path here
                Files.copy(file, toDirectory);
                return FileVisitResult.CONTINUE;
            }

            @Override
            public FileVisitResult preVisitDirectory(Path dir, BasicFileAttributes attrs) 
                    throws IOException {
                // In a full implementation, you'd need to create each 
                // sub-directory of the destination directory before 
                // copying files into it
                return super.preVisitDirectory(dir, attrs);
            }
        });
    }
}
于 2016-05-24T12:25:57.540 回答
11

由于 中的条件while，循环可能永远不会中断：

while (entry != null) {
  // If entry never becomes null here, loop will never break.
}

而不是null那里的检查，你可以试试这个：

ZipEntry entry = null;
while ((entry = zip.getNextEntry()) != null) {
  // Rest of your code
}
于 2013-03-27T19:00:02.077 回答
3

您可以使用示例代码让 Tika 为您处理容器文件。 http://wiki.apache.org/tika/RecursiveMetadata

据我所知，接受的解决方案不适用于嵌套 zip 文件的情况。然而，Tika 也会处理这种情况。

于 2013-12-24T22:38:57.187 回答
2

我实现这一点的方法是创建 ZipInputStream 包装类，该类将处理仅提供当前条目的流：

包装类：

public class ZippedFileInputStream extends InputStream {

    private ZipInputStream is;

    public ZippedFileInputStream(ZipInputStream is){
        this.is = is;
    }

    @Override
    public int read() throws IOException {
        return is.read();
    }

    @Override
    public void close() throws IOException {
        is.closeEntry();
    }

}

它的用途：

    ZipInputStream zipInputStream = new ZipInputStream(new FileInputStream("SomeFile.zip"));

    while((entry = zipInputStream.getNextEntry())!= null) {

     ZippedFileInputStream archivedFileInputStream = new ZippedFileInputStream(zipInputStream);

     //... perform whatever logic you want here with ZippedFileInputStream 

     // note that this will only close the current entry stream and not the ZipInputStream
     archivedFileInputStream.close();

    }
    zipInputStream.close();

这种方法的一个优点是：InputStreams 作为参数传递给处理它们的方法，并且这些方法倾向于在处理完输入流后立即关闭它。

于 2016-02-19T16:46:07.437 回答
0

我这样做了，记得更改 url 或 zip 文件 jdk 15

import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.net.MalformedURLException;
import java.net.URL;
import java.util.Scanner;
import java.util.stream.Stream;
import java.util.zip.ZipEntry;
import java.util.zip.ZipFile;
import java.io.*;
import java.util.*;
import java.nio.file.Paths;

class Main {
  public static void main(String[] args) throws MalformedURLException,FileNotFoundException,IOException{
    String url,kfile;
    Scanner getkw = new Scanner(System.in);
    System.out.println(" Please Paste Url ::");
    url = getkw.nextLine();
    System.out.println("Please enter name of file you want to save as :: ");
    kfile = getkw.nextLine();
    getkw.close();
    Main Dinit = new Main();
    System.out.println(Dinit.dloader(url, kfile));
    ZipFile Vanilla = new ZipFile(new File("Vanilla.zip"));
    Enumeration<? extends ZipEntry> entries = Vanilla.entries();

    while(entries.hasMoreElements()){
        ZipEntry entry = entries.nextElement();
//        String nextr =  entries.nextElement();
        InputStream stream = Vanilla.getInputStream(entry);
        FileInputStream inpure= new FileInputStream("Vanilla.zip");
        FileOutputStream outter = new FileOutputStream(new File(entry.toString()));
        outter.write(inpure.readAllBytes());
        outter.close();
    }

  }
  private String dloader(String kurl, String fname)throws IOException{
    String status ="";
    try {
      URL url = new URL("URL here");
      FileOutputStream out = new FileOutputStream(new File("Vanilla.zip"));         // Output File
      out.write(url.openStream().readAllBytes());
      out.close();
    } catch (MalformedURLException e) {
      status = "Status: MalformedURLException Occured";
    }catch (IOException e) {
      status = "Status: IOexception Occured";
    }finally{
      status = "Status: Good";}
    String path="\\tkwgter5834\\";
    extractor(fname,"tkwgter5834",path);
    

    return status;
  }
  private String extractor(String fname,String dir,String path){
    File folder = new File(dir);
    if(!folder.exists()){
      folder.mkdir();
    }
    return "";
  }
}
于 2021-04-06T18:37:04.157 回答