2

我正在尝试制作一个可以打开 docx 文件以读取、编辑和保存它的 Android 应用程序。

我的想法是将存档中的所有 xml 文件提取到临时文件夹。在这个文件夹中，我们可以编辑 .docx 中的内容/word/document.xml。问题是当我压缩这个临时文件夹以创建一个新的 docx 文件并替换旧文件时，在新的 docx 存档中，路径就像/mnt/sdcard/temp/"all files xml go here"xml 文件应该在第一级一样。

有人可以帮我解决这个问题吗？这是压缩临时目录的方法

注意：我使用的 dir2zip 参数的值为/mnt/sdcard/temp/***.docx

public void zipDir(String dir2zip, ZipOutputStream zos)
{
    try
   {
        //create a new File object based on the directory we
        //have to zip File   
        File zipDir = new File(dir2zip);

        //get a listing of the directory content
        String[] dirList = zipDir.list();
        byte[] readBuffer = new byte[2156];
        int bytesIn = 0;

        //loop through dirList, and zip the files
        for(int i=0; i<dirList.length; i++)
        {
            File f = new File(zipDir, dirList[i]);
            if(f.isDirectory())
            {
                    //if the File object is a directory, call this
                    //function again to add its content recursively
                String filePath = f.getPath();
                zipDir(filePath, zos);
                    //loop again
                continue;
            }
            //if we reached here, the File object f was not a directory
            //create a FileInputStream on top of f
            FileInputStream fis = new FileInputStream(f);
            //create a new zip entry
            ZipEntry anEntry = new ZipEntry(f.getPath());
            //place the zip entry in the ZipOutputStream object
            zos.putNextEntry(anEntry);
            //now write the content of the file to the ZipOutputStream
            while((bytesIn = fis.read(readBuffer)) != -1)
            {
                zos.write(readBuffer, 0, bytesIn);
            }
           //close the Stream
           fis.close();
        }
    }
    catch(Exception e)
    {
        //handle exception
    }
}
4

2 回答 2

0

我已经设法自己解决了。问题出在这一行：

    File f = new File(zipDir, dirList[i]);

它应该是

    File f = new File(dirList[i]);

如果包含参数 zipDir，则在存档中将使用目录的绝对路径！

于 2012-11-27T07:04:53.943 回答
0

通过进行以下两项修改，我现在设法使原始海报的代码在 Mac 和 Windows 上运行：

1：为每个目录添加一个ZipEntry：不要简单地忽略它

2：从ZipEntry名称中去掉目录名称

注意：zipinfo 很有用

这是一个适合我的程序：

import java.io.*;
import java.util.zip.*;

public class zipdoc
{
    String savedDir = null;

    public void zipDir(String dir2zip, ZipOutputStream zos)
    {
        try
       {
            if (savedDir == null)
                savedDir = dir2zip;
            // create a new File object based on the directory we
            // have to zip File   
            File zipDir = new File(dir2zip);

            //get a listing of the directory content
            String[] dirList = zipDir.list();
            byte[] readBuffer = new byte[2156];
            int bytesIn = 0;

            // loop through dirList, and zip the files
            for (int i=0; i<dirList.length; i++)
            {
                File f = new File(zipDir, dirList[i]);
                if (f.isDirectory())
                {
                    // if the File object is a directory, call this
                    // function again to add its content recursively
                    System.out.println("Adding dir: " + f); 
                    // create a new zip entry
                    ZipEntry anEntry = new ZipEntry(f.getPath().substring(savedDir.length()+1) + "/");
                    // place the zip entry in the ZipOutputStream object
                    zos.putNextEntry(anEntry);


                    String filePath = f.getPath();
                    zipDir(filePath, zos);
                    // loop again
                    continue;
                }
                else if (!f.getName().equals(".DS_Store"))
                {
                    // if we reached here, the File object f was not a directory
                    // and it's not the MacOSX special .DS_Store
                    // create a FileInputStream on top of f
                    System.out.println("Adding file: " + f);
                    FileInputStream fis = new FileInputStream(f);
                    // create a new zip entry
                    ZipEntry anEntry = new ZipEntry(f.getPath().substring(savedDir.length()+1));
                    // place the zip entry in the ZipOutputStream object
                    zos.putNextEntry(anEntry);
                    // now write the content of the file to the ZipOutputStream
                    while((bytesIn = fis.read(readBuffer)) != -1)
                    {
                        zos.write(readBuffer, 0, bytesIn);
                    }
                   // close the Stream
                   fis.close();
                }
            }
        }
        catch(Exception e)
        {
            // handle exception
            System.out.println(e);
        }
    }

    public void zipit(String inDir, String outFile)
    {
        try {
            ZipOutputStream zos = new ZipOutputStream(new FileOutputStream(new File(outFile)));
            zos.setMethod(0);
            zos.setMethod(ZipOutputStream.DEFLATED);
            zos.setLevel(0);

            zipDir(inDir, zos);
            zos.finish();
            zos.close();
        }
        catch (Exception e)
        {
            System.out.println(e);
        }
    }


    public static void main (String args[]) {
        zipdoc z1 = new zipdoc();

        // Check there are sufficient params if desired
        // first param is directory to be 'zipped', second is resulting
        // filename (??.docx)
        // eg java zipdoc dir1 newDoc.docx

        z1.zipit(args[0], args[1]);

        System.out.println("Finished creating " + args[1]);
    }
}
于 2013-01-10T10:04:17.343 回答