现在,I`ve下载的所有邮件,并将它们存储到
Message[] temp;
如何获取附件列表中为每个消息的
List<File> attachments;
注:没有第三方库,请,只是JavaMail的。
如果没有异常处理,但这里有云:
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);
}
}
问题是很老,但也许它会帮助别人。我想扩大大卫Rabinowitz`s答案。
if(!Part.ATTACHMENT.equalsIgnoreCase(bodyPart.getDisposition()))
像您期望不应返回所有附件,因为你可以将邮件,其中混合的部分是没有定义的配置。
----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())) {...}
编辑
有使用上述条件整个工作代码..因为每个部分可以封装另一个部件和附件应被嵌套在,递归用于通过所有部件遍历
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;
}
对于您保存附件文件中的代码部分的节省时间:
与使用javax邮件1.4版本后,你可以说
// 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();
你可以简单地使用Apache共享邮件API MimeMessageParser - getAttachmentList()一起共享IO和共享郎。
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
}
}
这里是我的mefi's solution的再现。
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);
}
}