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我真的有一个相当简单的问题。 我为供应商写了一个servlet来上传XML文件。 这些文件被写入到服务器上的某个位置。 所有文件都被重新命名为一个时间戳。Servlet文件并发
下面的代码是否存在并发问题的风险？ 我问，因为我们收到来自供应商的文件，看起来像 他们有2个不同的XML的文件
内容protected void doGet(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
processRequest(request, response);
}
protected void doPost(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
processRequest(request, response);
}
public String getServletInfo() {
return "Short description";
}// </editor-fold>
protected void processRequest(HttpServletRequest request,
HttpServletResponse response) throws ServletException, IOException {
File dirToUse;
boolean mountExists = this.getDirmount().exists();
if (!mountExists) {
this.log("MOUNT " + this.getDirmount() + " does not exist!");
dirToUse = this.getDiras400();
} else {
dirToUse = this.getDirmount();
}
boolean useSimpleRead = true;
if (request.getMethod().equalsIgnoreCase("POST")) {
useSimpleRead = !ServletFileUpload.isMultipartContent(request);
}
if (useSimpleRead) {
this.log("Handle simple request.");
handleSimpleRequest(request, response, dirToUse);
} else {
this.log("Handle Multpart Post request.");
handleMultipart(request, response, dirToUse);
}
}
protected void handleMultipart(HttpServletRequest request,
HttpServletResponse response, File dir) throws IOException,
ServletException {
try {
FileItemFactory fac = new DiskFileItemFactory();
ServletFileUpload upload = new ServletFileUpload(fac);
List<FileItem> items = upload.parseRequest(request);
if (items.isEmpty()) {
this.log("No content to read in request.");
throw new IOException("No content to read in request.");
}
boolean savedToDisk = true;
Iterator<FileItem> iter = items.iterator();
while (iter.hasNext()) {
FileItem item = (FileItem) iter.next();
getFilename(request);
File diskFile = new File(dir, this.getFilename(request));
item.write(diskFile);
if (!diskFile.exists()) {
savedToDisk = false;
}
}
if (!savedToDisk) {
throw new IOException("Data not saved to disk.");
}
} catch (FileUploadException fue) {
throw new ServletException(fue);
} catch (Exception e) {
throw new IOException(e.getMessage());
}
}
protected void handleSimpleRequest(HttpServletRequest request,
HttpServletResponse response, File dir) throws IOException {
// READINPUT DATA TO STRINGBUFFER
InputStream in = request.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
StringBuffer sb = new StringBuffer();
String line = reader.readLine();
while (line != null) {
sb.append(line + "\r\n");
line = reader.readLine();
}
if (sb.length() == 0) {
this.log("No content to read in request.");
throw new IOException("No content to read in request.");
}
//Get new Filename
String newFilename = getFilename(request);
File diskFile = new File(dir, newFilename);
saveDataToFile(sb, diskFile);
if (!diskFile.exists()) {
throw new IOException("Data not saved to disk.");
}
}
protected abstract String getFilename(HttpServletRequest request);
protected void saveDataToFile(StringBuffer sb, File diskFile) throws IOException {
BufferedWriter out = new BufferedWriter(new FileWriter(diskFile));
out.write(sb.toString());
out.flush();
out.close();
}
执行的getFileName：
@Override
protected String getFilename(HttpServletRequest request) {
Calendar current = new GregorianCalendar(TimeZone.getTimeZone("GMT+1"));
long currentTimeMillis = current.getTimeInMillis();
System.out.println(currentTimeMillis);
return "disp_" + request.getRemoteHost() + "_" + currentTimeMillis + ".xml";
}
无论如何，在此先感谢！
请问你的servlet维护任何状态（类级别的变量）？如果没有，我认为你没事。 – kosa 2012-02-14 16:03:29
感谢回复，是的，在类中的一些实例变量，但它们都是最终的...（目录信息，服务器IP地址，电子邮件地址）。 – Treurwilg 2012-02-14 16:07:20
记住，最终并不意味着对象状态不能改变，引用不能改变。只要你的类变量用于读取目的，我认为它是可以的。 – kosa 2012-02-14 16:30:50