Как сохранить вложения из MS Access 2007 с помощью Java?

1

У меня есть таблица в Access, где я храню id своей задачи и вложения (файл docx). Приложение имеет тип данных "Приложение". Мне нужно сохранить это приложение в дисковое пространство, используя java (каждый файл в собственном каталоге). Когда я экспортирую эту таблицу в XML, я вижу таких, как это

<Findings>
<id>265</id>
<Finding>
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AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
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AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
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AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
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AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA
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</FileData>
<FileName>Document.docx</FileName>
<FileType>docx</FileType>
</Finding>
<IDfind>PAY-14 1 07 01r7</IDfind>
</Findings>

с XML-тегами. Затем я использую DOM-парсер, анализируя этот XML и сохраняя в File (с расширением docx) каждый FileData. Но Word показать ошибку, когда я пытаюсь открыть этот файл. Можете ли вы помочь мне решить мою проблему? Вот мой код Java

File xmlFile = new File(filepath);
        try {
            DocumentBuilderFactory dbFactory = DocumentBuilderFactory
                    .newInstance();
            DocumentBuilder dBuilder = dbFactory.newDocumentBuilder();
            Document doc = dBuilder.parse(xmlFile);
            NodeList list = doc.getElementsByTagName("Findings");
            int length = list.getLength();
            Node nNode = null;
            for (int temp = 0; temp < length; temp++) {
                nNode = list.item(temp);
                if (nNode.getNodeType() == Node.ELEMENT_NODE) {
                    Element eElement = (Element) nNode;
                    String ids = eElement.getElementsByTagName("id").item(0)
                            .getTextContent();
                    Node finding = eElement.getElementsByTagName("Finding")
                            .item(0);
                    if (finding != null
                            && finding.getNodeType() == Node.ELEMENT_NODE) {
                        Element find = (Element) finding;
                        String findName = find.getElementsByTagName("FileName")
                                .item(0).getTextContent();
                        String data = find.getElementsByTagName("FileData")
                                .item(0).getTextContent();
                        String decdata = StringUtils.newStringUtf8(Base64
                                .decodeBase64(data));
                        String todec =  decdata.substring(decdata.indexOf("PK"));
                        String[] id = ids.split(" ");
                        if (id.length > 1) {
                            for (int idf = 0; idf < id.length; idf++) {
                                int idfind = Integer.parseInt(id[idf]);
                                File tempf = new File(defaultFileDirectory
                                        + idfind + "\\" + findName);
                                FileUtils.writeStringToFile(tempf, todec,
                                        "UTF8");

                            }
                        } else {
                            int idf = Integer.parseInt(ids);
                            File tempf = new File(defaultFileDirectory + idf
                                    + "\\" + findName);
                            FileUtils.writeStringToFile(tempf, decdata, "UTF8");
                        }
                    }
} catch (Exception e) {
            e.printStackTrace();
        }
    }
Теги:
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1 ответ

0
Лучший ответ

Обычно бинарные данные кодируются Base64 при включении в файл XML. Я взял <FileData> из вашего примера, Base64 расшифровал его, а затем написал в файл. В результате получилось:

Изображение 174551

Это похоже на файл.docx с 22 байтами "мусора" в самом начале. (Файл.docx на самом деле является ZIP-архивом, поэтому первые два байта файла должны быть "PK" или 0x50 0x4B.)

Мне удалось успешно создать документ Word:

  • извлечение текста <FileData>,
  • Base64 декодирует его,
  • 0x50 0x4B ведущие байты, предшествующие "PK" (0x50 0x4B), и
  • сохраняя остальное как файл.docx.

Пример реализации. После того, как текст <FileData> извлечен в переменную String с именем strEncoded вы можете извлечь файл.docx, используя следующий код:

byte[] bytesDecoded = DatatypeConverter.parseBase64Binary(strEncoded);
int pkStart = -1;
for (int i = 0; i < (bytesDecoded.length - 1); i++) {
    if (bytesDecoded[i] == 0x50) {  // "P"
        if (bytesDecoded[i+1] == 0x4B) {  // "K"
            pkStart = i;
            break;
        }
    }
}
if (pkStart < 0) {
    System.out.println("Did not find 'PK' marker in decoded data.");
}
else {
    FileOutputStream fos = new FileOutputStream("C:\\Users\\Gord\\Desktop\\Hello.docx");
    fos.write(Arrays.copyOfRange(bytesDecoded, pkStart, bytesDecoded.length));
    fos.close();
}
Поделиться
  • 0
    Я пишу только небольшую часть «<FileData>». Теперь я редактирую свой вопрос и помещаю все данные
  • 0
    @Sydorov Мне удалось успешно извлечь документ, используя шаги, изложенные в моем ответе. После того как я Base64-декодировал текст <FileData> , удалил 22 байта «мусора» в начале и сохранил остальные как файл .docx, я мог открыть его без проблем в Word 2010.
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