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$
Related
I have created the default play application in IntelliJ in directory P. I have over-written the default index.scala.html with my own html code. The html code refers to some css and js files which are outside the directory P. To include these external files, I added the directory of these files using project configuration settings.
My webpage doesn't load properly as the server returns 404 for the css and js files. What am I doing wrong?
When you added your directory using project structure, you only say:
Hey, IDEA, please consider this folder part of my project, consider
its contents source code and display it when I open my project.
However, when you deploy or run your app, you only deploy the usual folders to the server, which contain the resources which will be available for clients to access.
The external directory is not part of these directories and will not be deployed.
What you can do is to copy the file from the external directory as a part of your build process before deploying the application.
EDIT: Detailed answer here: What is intellij's build process for play applications
i am making a 'simple' program that on button click does this:
Private Sub Button4_Click(sender As Object, e As EventArgs)
Dim proc As New System.Diagnostics.Process()
proc = Process.Start("resources\navcoder.exe", "")
proc.WaitForExit()
End Sub
all works fine when testing in visual studio but not once i publish and install, even if the resource folder is in the install directory.
if i change it to:
proc = Process.Start("c:\resources\navcoder.exe", "")
it works fine, but i obviuosly needs to have the folder there with the required files in it.
what am i doing wrong?
A lot of people seem not to realise what resources are. The whole point of adding a resource to your project is to have the data it contains compiled into your EXE. The Resources folder in your project is just a place to store the original source files. It doesn't exist as far as the application is concerned, just as your VB source code files don't exist. When you build your project, the data in those resources is compiled into your EXE so they are no longer files and can no longer be used as files.
That's why you don't embed other EXE files as resources. You could extract the resource and save it as a file first but I would recommend against that. Add a new folder to your project and add the EXE file and any dependencies to that folder and set their Build Action to Content. They will then be copied to your output folder as is. You can then execute the EXE file because it is an EXE file. You should also use Application.StartupPath as the root of the file path rather then relying on the current directory being what you think it will be.
You should always check for existance, if its a web app, would do this in the application start in the global asax. If you are running this under a different user account, check the account has permissions. Otherwise even if the directory exists, it will fail if it cant access it or does not have permissions to execute the file.
Might be able to tell you more if you give the actual exception. I.e Unauthorized or does not exist.
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
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.