JavaMail带附件时出错

JavaMail requires an InputStreamSource that creates a fresh stream for every

public void addAttachment(String attachmentFilename, DataSource dataSource) throws MessagingException {
Assert.notNull(attachmentFilename, "Attachment filename must not be null");
Assert.notNull(dataSource, "DataSource must not be null");

try {
MimeBodyPart mimeBodyPart = new MimeBodyPart();
mimeBodyPart.setDisposition("attachment");
mimeBodyPart.setFileName(MimeUtility.encodeText(attachmentFilename));
mimeBodyPart.setDataHandler(new DataHandler(dataSource));
this.getRootMimeMultipart().addBodyPart(mimeBodyPart);
} catch (UnsupportedEncodingException var4) {
throw new MessagingException("Failed to encode attachment filename", var4);
}
}

public void addAttachment(String attachmentFilename, File file) throws MessagingException {
Assert.notNull(file, "File must not be null");
FileDataSource dataSource = new FileDataSource(file);
dataSource.setFileTypeMap(this.getFileTypeMap());
this.addAttachment(attachmentFilename, (DataSource)dataSource);
}

public void addAttachment(String attachmentFilename, InputStreamSource inputStreamSource) throws MessagingException {
String contentType = this.getFileTypeMap().getContentType(attachmentFilename);
this.addAttachment(attachmentFilename, inputStreamSource, contentType);
}

public void addAttachment(String attachmentFilename, InputStreamSource inputStreamSource, String contentType) throws MessagingException {
Assert.notNull(inputStreamSource, "InputStreamSource must not be null");
if (inputStreamSource instanceof Resource && ((Resource)inputStreamSource).isOpen()) {
throw new IllegalArgumentException("Passed-in Resource contains an open stream: invalid argument. JavaMail requires an InputStreamSource that creates
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a fresh stream for every call.");
} else {
DataSource dataSource = this.createDataSource(inputStreamSource, contentType, attachmentFilename);
this.addAttachment(attachmentFilename, dataSource);
}
}

InputStreamResource inputStreamResource

messageHelper.addAttachment(MimeUtility.encodeWord(fileName), new ByteArrayResource(IOUtils.toByteArray(inputStreamResource.getInputStream())));