I'm new to Cloudbees. I just opened an account and trying to upload a .JAR file which basically downloads a file to the location mentioned by user (through java command line arguments). So far I have run the .JAR in my local. So far, I was referring to my local file system to save the file. If I deploy my .JAR file to Cloudbees SDK, where can I save the downloaded file (and then process it).
NOTE: I know this is not a new requirement in java if we deploy the jar in UNIX/WINDOWS OS where we can refer the file system w.r.t to home directory.
Edit#1:
I've seen couple of discussions about the same topic.
link#1
link#2
Everywhere they are talking about the ephemeral (temporary) file system which we can access through System.getProperty("java.io.tempDir"). But I'm getting null when I access java.io.tempDir. Did anyone manage to save files temporarily using this tempDir?
You can upload a jar with the java stack specifying the class and classpath (http://developer.cloudbees.com/bin/view/RUN/Java+Container)
Our filesystem however is not persistent, so if you are talking about saving a file from within your application, you could save it in this path
System.getProperty("java.io.tmpdir")
but it will be gone when your application hibernates, scales-up/down or is redeployed.
If you want a persistent way to store file/images, you can use AmazonS3 from your cloudbees application: uploading your files there will ensure their persistence.
You can find an example of how to do that in this clickstart:
http://developer-blog.cloudbees.com/search?q=amazon+s3
More information here: https://wiki.cloudbees.com/bin/view/RUN/File+system+access
Related
I am having problems with accessing file in public directory after deployment.
The stage command give me a folder target\universal\bin where my .exe file is.
In development mode I used to upload my files to public\uploads\pictures and access them from this location. But, after deployment I am unable to upload the pictures. I read this Stack Link that has two options. Is it possible to define a folder directory that is not absolute.
Application Conf
myUploadPath="public/Upload/Pictures/"
Accessing folder
String myUploadPath = Play.application().configuration()
.getString(myUploadPath);
Please tell me a solution to overcome this..
Found a solution
During development we are using public directory to store anything extras(in my case uploaded files) that we have. But, while deploying the application it is important that we change the reference to these extra files. I have changed the path from public directory to the absolute path of where the file is executed using
Play.application().path().getAbsolutePath()
and store the files in this directory.
If you don't want to use this than you can also specify an External Asset
I am making a Java application that uses, Spring, Maven and the AWS-SDK-Java. In order to the AWS SDK to work I have to place the AWSCredentials.properties file inside the "MyProject/src/main/resources" folder.
So far so so good. Then I have to create a .war file. To do that I use mvn install command and voilá.
Now inside the .war created, the file I want to access is in the "/WEB-INF/classes/" folder and I need to read it.
How can I access this file inside the war so I can read its content? I have experimented with ServeltContext but so far nothing I try works!
It is generally not a good practice to keep credential in code package (war). Instead I would suggest you use IAM Roles.
This will make it easy for you to move your code from one AWS account to another (say dev environment to production). Since the code will be submitted to a version control system which will be accessed by many, it is also good from a security point of view to use IAM roles.
I found a way to do it. I can access the file by using:
InputStream inputStream =
getClass().getClassLoader().getResourceAsStream("/resources/email/file.txt");
As explained in this discusison:
Reading a text file in war archive
I am doing a program to file-uplaod in the openshift. In this program I am upload whole files documents in the uploads folder, which is given default folder of the Openshift server.
I am using the file upload path
System.getenv("OPENSHIFT_DATA_DIR")
for upload files.
But I want to upload the file in my own project folder. I am created folder externally and I want to all the file in this folder.
$OPENSHIFT_DATA_DIR is the correct location for persistent data. If you upload to your $OPENSHIFT_REPO_DIR, subsequent git pushes will overwrite contents of with your local repo and the uploads will be lost.
If I'm misunderstanding your use case, please help me understand the reference to 'project folder' and 'created folder externally'. Feel free to share your use case on the OpenShift Forums as well: https://www.openshift.com/forums/openshift
I am writing an application (basically eclipse plugin), so there are few combo-box, drop-downs etc, which I am getting values for them dynamically via XML file. My doubt is which is the best way to store these files in a particular directory so that it can be read in both Windows as well as Linux.
Initially I tried to create a config path under {eclipse.home.location} like:
String finalPath = System.getProperty("eclipse.home.location") +"/myAppConfig";
and store all of my plugin's configuration there (not only configuration but few helper jars which I programatically read in my plugin). But there is a probabilty that Eclipse installation maybe in shared location and user may not have write access to it.
I also tried to store it in a program files directory using:
System.getenv("ProgramFiles");
But this fails under non-windows environments. So my question is can anyone shed a light on this so that I can store in some common directory where it is valid for both windows and linux?
Kindly let me know if my wordings are confusing. Or is it possible to store my config files under plugins directory and get the path like this /plugins/myConfigDir ?
Try using the getStateLocation() method in Plugin.
That will give you an IPath that points to a user specific workspace location.
How can I programmatically find the path of JNLP file?I am using Java Web Start to generate the JNLP file.
I know that manually you can find the JNLP file in the Java Cache Viewer in Resources with the name launch.jnlp, but I really need to know if there exists a Java class that can programmatically find the jnlp file by searching the in-memory cache.
How can I programatically find the path of JNLP file?
It is hidden deliberately. If you can find it from within code (shorting of asking the user to browse to it), the JRE has a security bug.
..for launching the installer I need the application jar files path
No you don't.
Put the installer inside a Jar.
Add the Jar to the resources of the app.
Get an URL using getResource(String).
Read the byte[] and write it to a temporary file at a known location (e.g. java.io.tmpdir).
Launch the installer from the known location.
String jnlpPath = System.getProperty("jnlpx.origFilenameArg");