Servlet 文件并发
我真的有一个很简单的问题。 我为供应商编写了一个 servlet,用于上传 XML 文件。 这些文件被写入服务器上的某个位置。 所有文件都使用时间戳进行重命名。
下面的代码是否存在并发问题的风险? 我问是因为我们从供应商那里收到文件,看起来像 他们有来自 2 个不同 XML 文件
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();
}
getFileName 实现的内容:
@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,提前感谢!
I have quite a simple question really.
I wrote a servlet for suppliers to upload XML-files to.
These files get written to a location on the server.
All the files get renamed with a timestamp.
Is there a risk of concurrency problems with the code below?
I ask because we receive files from a supplier, that look like
they have content from 2 different XML-files
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();
}
getFileName implementation:
@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";
}
Anyway, thanks in advance!
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不会出现同步问题,但可能会出现竞争条件,例如,两个线程可能使用 getFileName() 方法返回相同的文件名
There would not be synchronization problems but there can be race conditions, for example, two threads might return the same file name using the method getFileName()