how to judge if the file is doc or docx in POI

问题:

The title may be a little confusing. The simplest method must be judging by extension name just like:

// is represents the InputStream   
if (filePath.endsWith("doc")) {
    WordExtractor ex = new WordExtractor(is);
    text = ex.getText();
    ex.close();
} else if(filePath.endsWith("docx")) {
    XWPFDocument doc = new XWPFDocument(is);
    XWPFWordExtractor extractor = new XWPFWordExtractor(doc);
    text = extractor.getText();
    extractor.close();
}

This works in most cases. But I have found that for certain file whose extension is doc (a docx file essentially) if you open using winrar, you will find xml files. As it is known that a docx file is a zip file consists of xml files. I believe this problem must not be rare. But I have not found any information about this. Obviously, judging by extension name to read a doc or docx is not appropriate.

In my case, I have to read a lot of files. And I will even read the doc or docx inside a compressed file, zip, 7z or even rar. Hence, I have to read content by inputStream instead of a File or something else. So how to know whether a file is .docx or .doc format from Apache POI is totally not suitable for my case with ZipInputStream.

What is the best way to judge a file is a doc or docx? I want a solution to read the content from a file which may be doc or docx. But not only just simply judge if it is a doc or docx. Apparently, ZipInpuStream is not a good method for my case. And I believe it is not a appropriate method for others either. Why do I have to judge if the file is doc or docx by an exception?

回答1:

Using the current stable apache poi version 3.17 you may use FileMagic. But internally this will of course also have a look into the files.

Example:

import java.io.InputStream;
import java.io.FileInputStream;
import java.io.BufferedInputStream;

import org.apache.poi.poifs.filesystem.FileMagic;

import org.apache.poi.hwpf.extractor.WordExtractor;
import org.apache.poi.xwpf.extractor.XWPFWordExtractor;
import org.apache.poi.xwpf.usermodel.XWPFDocument;

public class ReadWord {

 static String read(InputStream is) throws Exception {

System.out.println(FileMagic.valueOf(is));

  String text = "";

  if (FileMagic.valueOf(is) == FileMagic.OLE2) {
   WordExtractor ex = new WordExtractor(is);
   text = ex.getText();
   ex.close();
  } else if(FileMagic.valueOf(is) == FileMagic.OOXML) {
   XWPFDocument doc = new XWPFDocument(is);
   XWPFWordExtractor extractor = new XWPFWordExtractor(doc);
   text = extractor.getText();
   extractor.close();
  }

  return text;

 }

 public static void main(String[] args) throws Exception {

  InputStream is = new BufferedInputStream(new FileInputStream("ExampleOLE.doc")); //really a binary OLE2 Word file
  System.out.println(read(is));
  is.close();

  is = new BufferedInputStream(new FileInputStream("ExampleOOXML.doc")); //a OOXML Word file named *.doc
  System.out.println(read(is));
  is.close();

  is = new BufferedInputStream(new FileInputStream("ExampleOOXML.docx")); //really a OOXML Word file
  System.out.println(read(is));
  is.close();

 }
}

回答2:

try {
    new ZipFile(new File("/Users/giang/Documents/a.doc"));
    System.out.println("this file is .docx");
} catch (ZipException e) {
    System.out.println("this file is not .docx");
    e.printStackTrace();
}