现在我已经下载了所有的消息,并将它们存储到
Message[] temp;
如何获取每封邮件的附件列表
List<File> attachments;
注意:没有第三方库,请只使用JavaMail。
发布于 2009-11-17 19:30:11
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);
}
}
发布于 2013-08-14 17:49:32
这个问题已经很古老了,但也许它会对某些人有所帮助。我想扩展一下David Rabinowitz`s的答案。
if(!Part.ATTACHMENT.equalsIgnoreCase(bodyPart.getDisposition()))
不应该像你期望的那样返回所有附件,因为你可能会有混合部分没有定义的处理的邮件。
----boundary_328630_1e15ac03-e817-4763-af99-d4b23cfdb600
Content-Type: application/octet-stream;
name="00000000009661222736_236225959_20130731-7.txt"
Content-Transfer-Encoding: base64
因此,在这种情况下,您还可以检查文件名。如下所示:
if (!Part.ATTACHMENT.equalsIgnoreCase(part.getDisposition()) && StringUtils.isBlank(part.getFileName())) {...}
编辑
上面描述了使用上述条件的完整工作代码。因为每个部件都可以封装另一个部件,并且应该嵌套附件,所以使用递归遍历所有部件
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;
}
发布于 2014-07-04 22:16:08
保存附件文件的代码的一些节省时间的方法:
使用javax mail version 1.4及更高版本,您可以这样说
// 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();
https://stackoverflow.com/questions/1748183
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