java - 使用 Java Mail 下载附件

Question

现在我已经下载了所有的消息，并将它们存储到

Message[] temp;

如何获取这些邮件的附件列表

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，但这里有：

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

问题很老了，但也许它会对某人有所帮助。我想扩展大卫拉比诺维茨的答案。

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

不应该像您期望的那样返回所有附件，因为您可以收到混合部分没有定义处置的邮件。

   ----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())) {...}

编辑

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应该嵌套附件，所以使用递归遍历所有部分

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

保存附件文件的代码可以节省一些时间：

使用 javax 邮件版本1.4及更高版本，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());

代替

    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

您可以简单地使用 Apache Commons Mail API MimeMessageParser - getAttachmentList()以及 Commons IO 和 Commons Lang。

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

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
    }