我需要在SFTP文件夹中获取文件名的递归列表。现在我使用以下代码:
private void getFilesRecursive(List<LsEntry> fileList, List<FileDto> response, String dirParentName,
SftpManagerWithPool manager) throws SftpException {
for (LsEntry file : fileList) {
if (!file.getAttrs().isDir()) {
response.add(new FileDto(file.getFilename(), StorageType.FOLDER.getName(),
new Attributes(file.getAttrs().getATime(), file.getAttrs().getMTime(),
file.getAttrs().getAtimeString(), file.getAttrs().getMtimeString(),
file.getAttrs().getPermissionsString(), getPathWitoutRootDirectoryt(dirParentName),
file.getAttrs().getSize(), file.getAttrs().isDir())));
} else if (file.getAttrs().isDir() && !".".equals(file.getFilename())) {
List<LsEntry> files = manager.listFiles(context.getBasePath().concat("/")
.concat(dirParentName.concat("/").concat(file.getFilename())));
getFilesRecursive(files, response, dirParentName.concat("/").concat(file.getFilename()), manager);
}
}
}
public List<FileDto> getFiles() throws ServiceException {
Session session = null;
SftpManagerWithPool manager = null;
try {
session = getSession();
manager = new SftpManagerWithPool(session);
manager.connect();
List<FileDto> response = new ArrayList<>();
List<LsEntry> files = manager
.listFiles(context.getBasePath().concat("/").concat(context.getPathToProcess()));
getFilesRecursive(files, response, context.getPathToProcess(), manager);
return response;
} catch (Exception e) {
throw new ServiceException(HttpStatus.INTERNAL_SERVER_ERROR.value(), e.getMessage(),
e.getLocalizedMessage());
} finally {
if (manager != null) {
manager.disconnect();
try {
returnSession(session);
} catch (Exception e) {
throw new ServiceException(HttpStatus.INTERNAL_SERVER_ERROR.value(), e.getMessage(),
e.getLocalizedMessage());
}
}
}
}
以下是SftpManagerWithPool中的方法:
@SuppressWarnings("unchecked")
public List<LsEntry> listFiles(String pathFrom) throws SftpException {
if (c == null || session == null || !session.isConnected() || !c.isConnected()) {
throw new SftpException(ErrorCode.LIST_FILES.ordinal(), "Connection to server is closed. Open it first.");
}
String filePath = pathFrom + "/";
return c.ls(filePath);
}
一切都运行良好,不超过5k文件,但是当我们有更多文件时,需要花费大量时间并导致超时。
我的问题是,如何使用c.ls(filePath);
方法仅列出文件夹中的前N个文件?我正在寻找类似于Linux shell命令ls -U | head -4
---- ----- EDIT
我修改了我的方法listFiles
如下,但我仍然没有获得更好的表现:
public List<LsEntry> listFiles(String pathFrom, int top) throws SftpException {
if (c == null || session == null || !session.isConnected() || !c.isConnected()) {
throw new SftpException(ErrorCode.LIST_FILES.ordinal(), "Connection to server is closed. Open it first.");
}
String filePath = pathFrom + "/";
List<LsEntry> response = new ArrayList<>();
c.ls(filePath, new LsEntrySelector() {
@Override
public int select(LsEntry record) {
if (response.size() <= top) {
response.add(record);
return LsEntrySelector.CONTINUE;
} else {
return LsEntrySelector.BREAK;
}
}
});
return response;
}
非常感谢你:)
答案 0 :(得分:2)
使用overload of ChannelSftp.ls
method that takes LsEntrySelector
interface:
public void ls(String path, LsEntrySelector selector)
实施LsEntrySelector.select
以收集文件条目。一旦你有足够的,让它返回BREAK
。