I am using liferay 6.0.5. I am uploading file using following code.
UploadPortletRequest uploadRequest = PortalUtil.getUploadPortletRequest(actionRequest);
String submissionFileName = uploadRequest.getFileName("file");//uploaded filename
File submissionFile = uploadRequest.getFile("file");
this works fine and upload file in tomcat's temp directory with some different name. What I want is.."There is one folder docs in my project directory. I want uploaded file in this directory". How to do this in liferay..?
If you want to change the temp directory for the file upload then you can change the following property in portal-ext.properties:
com.liferay.portal.upload.UploadServletRequestImpl.temp.dir=C:/MyTempDir
Hope this is what you are looking for.
If you want to add files to Liferay's Document Library then check out the following classes:
http://docs.liferay.com/portal/6.0/javadocs/com/liferay/portlet/documentlibrary/service/DLFileEntryLocalServiceUtil.html
http://docs.liferay.com/portal/6.0/javadocs/com/liferay/portlet/documentlibrary/service/DLFolderLocalServiceUtil.html
Related
I created "xlsfiles" subfolder in public folder, and I used it folder for create xls file by request.
And when user send request to server to get this xls file I serve this file to client.
When I run project on Dev Mode all working perfectly, but when I run project on Production Mode not exist "xlsfiles" subfolder and I get "FileNotFoundException" exception.
Have any solution? How to make create sub folder in the project and create file in it?
Use Files class from standard Play framework API:
http://www.playframework.com/documentation/2.2.x/api/scala/index.html#play.api.libs.Files$
I am properly sending a nsurlrequest to download a known path file and then save it to document directory. However, now I need to download all files that a remote directory contains. I know path for directory but not which files are inside. How to list that files in order to build paths to download it? Thank you.
Unfortunately unless you parsed a index file containing a list of the files, this is not possible as the HTTP protocol does not support directory listing. You would have to use an FTP server instead
How to upload file, create new folder using ssh_connent function php.
get.php file uploaded http://xxxx.com/demo/review/
But the review folder doesn't exist on the xxxx.com. I want to create review folder then upload the get.php file.
See ssh2_ftp function manual:
http://php.net/manual/en/function.ssh2-sftp.php
I have added some files that I need to be downloaded to the Application start up path. So I set Build Action as content now the files have been copied some where
C:\Documents and Settings\TestUser.ANNAM\Local Settings\Apps\2.0\Data\HVDRBMY5.8AA\858AT9VM.TNP\test..tion_2d7cfc137d9c2c74_0001.0013_432bd4561850d290\Data
How can access file from the application. My problem since it is a dynamic path will it be same folder count so that we can use like ....\Data\ Some think like this
You can use My.Application.Info.DirectoryPath this will return the directory where the application is stored.
I have added some files that I need to be downloaded to the Application start up path. So I set Build Action as content now the files have been copied some where
C:\Documents and Settings\TestUser.ANNAM\Local Settings\Apps\2.0\Data\HVDRBMY5.8AA\858AT9VM.TNP\test..tion_2d7cfc137d9c2c74_0001.0013_432bd4561850d290\Data
How can access file from the application. My problem since it is a dynamic path will it be same folder count so that we can use like ..\..\Data\ Some think like this
Application.UserAppDataPath gets the path for the application data of a user.
Application.StartupPath gives you the path for the executable file that started the application, not including the executable name.
Starting with one of these, you should be able to use System.IO to manipulate the paths until you get the folder where your data files are.