Question

现在我已经下载了所有消息，并将它们存储到

Message[] temp;

如何获取每条消息的附件列表

List<File> attachments;

注意：请不要使用第三方库，只需JavaMail。

Answer 1

没有异常处理，但这里是：

List<File> attachments = new ArrayList<File>();
for (Message message : temp) {
    Multipart multipart = (Multipart) message.getContent();

    for (int i = 0; i < multipart.getCount(); i++) {
        BodyPart bodyPart = multipart.getBodyPart(i);
        if(!Part.ATTACHMENT.equalsIgnoreCase(bodyPart.getDisposition()) &&
               StringUtils.isBlank(bodyPart.getFileName())) {
            continue; // dealing with attachments only
        } 
        InputStream is = bodyPart.getInputStream();
        // -- EDIT -- SECURITY ISSUE --
        // do not do this in production code -- a malicious email can easily contain this filename: "../etc/passwd", or any other path: They can overwrite _ANY_ file on the system that this code has write access to!
//      File f = new File("/tmp/" + bodyPart.getFileName());
        FileOutputStream fos = new FileOutputStream(f);
        byte[] buf = new byte[4096];
        int bytesRead;
        while((bytesRead = is.read(buf))!=-1) {
            fos.write(buf, 0, bytesRead);
        }
        fos.close();
        attachments.add(f);
    }
}

Answer 2

问题很老，但也许会对某人有所帮助。我想扩展David Rabinowitz的答案。

if(!Part.ATTACHMENT.equalsIgnoreCase(bodyPart.getDisposition()))

不应该按照您的预期返回所有内容，因为您可以拥有邮件，其中混合部分没有定义的处置。

   ----boundary_328630_1e15ac03-e817-4763-af99-d4b23cfdb600
Content-Type: application/octet-stream;
    name="00000000009661222736_236225959_20130731-7.txt"
Content-Transfer-Encoding: base64

所以在这种情况下，您还可以检查文件名。像这样：

if (!Part.ATTACHMENT.equalsIgnoreCase(part.getDisposition()) && StringUtils.isBlank(part.getFileName())) {...}

修改

使用上面描述的条件有完整的工作代码。因为每个部分都可以封装另一个部分并且附件应该嵌套，所以递归用于遍历所有部分

public List<InputStream> getAttachments(Message message) throws Exception { Object content = message.getContent(); if (content instanceof String) return null; if (content instanceof Multipart) { Multipart multipart = (Multipart) content; List<InputStream> result = new ArrayList<InputStream>(); for (int i = 0; i < multipart.getCount(); i++) { result.addAll(getAttachments(multipart.getBodyPart(i))); } return result; } return null; } private List<InputStream> getAttachments(BodyPart part) throws Exception { List<InputStream> result = new ArrayList<InputStream>(); Object content = part.getContent(); if (content instanceof InputStream || content instanceof String) { if (Part.ATTACHMENT.equalsIgnoreCase(part.getDisposition()) || StringUtils.isNotBlank(part.getFileName())) { result.add(part.getInputStream()); return result; } else { return new ArrayList<InputStream>(); } } if (content instanceof Multipart) { Multipart multipart = (Multipart) content; for (int i = 0; i < multipart.getCount(); i++) { BodyPart bodyPart = multipart.getBodyPart(i); result.addAll(getAttachments(bodyPart)); } } return result; }

Answer 3

为保存附件文件的代码节省了一些时间：

使用javax邮件版本 1.4 之后，您可以说

// SECURITY LEAK - do not do this! Do not trust the 'getFileName' input. Imagine it is: "../etc/passwd", for example.
// bodyPart.saveFile("/tmp/" + bodyPart.getFileName());

而不是

    InputStream is = bodyPart.getInputStream();
    File f = new File("/tmp/" + bodyPart.getFileName());
    FileOutputStream fos = new FileOutputStream(f);
    byte[] buf = new byte[4096];
    int bytesRead;
    while((bytesRead = is.read(buf))!=-1) {
        fos.write(buf, 0, bytesRead);
    }
    fos.close();

Answer 4

您可以在Commons IO和Commons Lang中使用Apache Commons Mail API MimeMessageParser - getAttachmentList()。

MimeMessageParser parser = ....
parser.parse();
for(DataSource dataSource : parser.getAttachmentList()) {

    if (StringUtils.isNotBlank(dataSource.getName())) {}

        //use apache commons IOUtils to save attachments
        IOUtils.copy(dataSource.getInputStream(), ..dataSource.getName()...)
    } else {
        //handle how you would want attachments without file names
        //ex. mails within emails have no file name
    }
}

Answer 5

以下是我mefi's solution的演绎。

private static void attachments(
    final BodyPart body, final BiConsumer<String, InputStream> consumer)
    throws MessagingException, IOException {
    final Multipart content;
    try {
        content = (Multipart) body.getContent();
        for (int i = 0; i < content.getCount(); i++) {
            attachments(content.getBodyPart(i), consumer);
        }
        return;
    } catch (final ClassCastException cce) {
    }
    if (!Part.ATTACHMENT.equalsIgnoreCase(body.getDisposition())) {
        return;
    }
    final String name = body.getFileName();
    if (name == null || name.trim().isEmpty()) {
        return;
    }
    try (final InputStream stream = body.getInputStream()) {
        consumer.accept(name, stream);
    }
}

public static void attachments(
    final Message message, final BiConsumer<String, InputStream> consumer)
    throws IOException, MessagingException {
    final Multipart content;
    try {
        content = (Multipart) message.getContent();
    } catch (final ClassCastException cce) {
        return;
    }
    for (int i = 0; i < content.getCount(); i++) {
        attachments(content.getBodyPart(i), consumer);
    }
}

Answer 6

返回带有附件的身体部位列表。

@Throws(Exception::class)
    fun getAttachments(message: Message): List<BodyPart>{
        val content = message.content
        if (content is String) return ArrayList<BodyPart>()
        if (content is Multipart) {
            val result: MutableList<BodyPart> = ArrayList<BodyPart>()
            for (i in 0 until content.count) {
                result.addAll(getAttachments(content.getBodyPart(i)))
            }
            return result
        }
        return ArrayList<BodyPart>()
    }

    @Throws(Exception::class)
    private fun getAttachments(part: BodyPart): List<BodyPart> {
        val result: MutableList<BodyPart> = ArrayList<BodyPart>()

        if (Part.ATTACHMENT == part.disposition && !part.fileName.isNullOrBlank()){
            result.add(part)
        }

        val content = part.content
        if (content is Multipart) {
            for (i in 0 until (content ).count) {
                val bodyPart = content.getBodyPart(i)
                result.addAll(getAttachments(bodyPart)!!)
            }
        }
        return result
    }

使用Java Mail下载附件

6 个答案: