Develop Reference

How to avoid hard code full path in Paths#get method

I want to save a copy of Primefaces UploadedFile to my project directory. I have been searching the internet for the solution, what I have found is using Paths#get method. The example given in this answer is Paths.get("/path/to/uploads");, the problem is, where is the exact path of /path/to/uploads? I can't find it. I mean where should I create the path /path/to/uploads? Under my project directory? but which folder? I solve this issue temporary by hard coding the full path like Paths.get("C:/uploads/");

FacesContext.getCurrentInstance().getExternalContext().getRealPath("/") will return you the current installation directory of your project.
And as #Kukeltje suggested, never ever save an uploaded file to your project directory, ... save it outside the webapps or even outside your container.
Therefore create a directory outsite your container (where you want to place your uploaded copies) and append ../ to the above path for each back step.
Say, if your application is deployed at D:/Tools/Tomcat7/webapps/your-application-name (e.g. on Windows using Tomcat) and you want to save copies to D:/Tools/uploads then following will give you required file path:
String uploadsFilePath = FacesContext.getCurrentInstance().getExternalContext()
.getRealPath("../../../uploads");
Use it with the Paths.get(uploadsFilePath) and develop your download logic (I am not sure which library you are using for the Paths class).

How about getClassLoader().getResource(Path/to/file)
So like
MyClass.class.getResource(bla/bla)
Which are now nested in src/resources
Like this You are system independent
#profit

You have several options:
For very quick development, you can use a hardcoded path, but be sure that it exists in your SUT (system under test).
You can define it as a static final string in your module, but this means that each time you want to change that path, you will need to recompile...
You can read that value from a property/config file.
There are more options (like using the registry if you are on Windows, or using an environment variable).

how to define a file path in java to make it ready for production phase?

I am writing a program in java with netbeans IDE which receives a jasper report *.jrxml and then displays the report for the user. I wrote the following line of code for the file path
String reportSource = "src\\jasper-reports\\report.jrxml";
but when I move the dist folder in some other place and try to run the jar file inside it, my program can not find the report.
my problem is that were should I put the *.jrxml file and how to define it's path in my program so that when I want to give my software to someone else it runs without any errors (e.g. the program can find the file)

avoid using absolute paths. try to include the file as a resource in your netbeans project. then in your code you can search for and load the file as
new InputStreamReader((Main.class.getResourceAsStream("/report.jrxml")))
something like that depending on where the file resides in your project

it's more recommended using one of the two approaches:
pass the locations/paths as a -Dproperty=value in the Java application launcher command line http://www.tutorialspoint.com/unix_commands/java.htm
store it the locations/paths in a configurations file with a unique key and edit the file accordingly for different environments,
e.g.this files always stored in ${HOME}/config_files/ directory
absolute paths considered a bad practice

How to give relative path for my file in java file?

I am using tomcat server for my java application(Servlet,jsp).In my servlet page calling one java class function.Actually the java file is written separately and i will call the function written inside it.With in the function, i have to read one config(user defined) file for some purpose.so, i am using File class for that.
But, here i have to give relative path of the config file(user defined).Because, now i am running this application in local windows server.But my live server is based on Linux.So, the file path is changed in linux.
File f1=new File("D:\tomcat\webapp\myapp\WEB-INF\src\point_config.txt"); -- windows
File f1=new File("D:\ravi\tomcat\webapp\myapp\WEB-INF\src\point_config.txt"); -- linux
So, i have to give relative path of the file that is common to both windows and linux machine.
Is there a way to do this?
Please guide me to get out of this issue?

Place your config file under your webapp WEB-INF/classes folder and read like this in code
InputStream is=
YourClassName.class.getResourceAsStream("point_config.txt");

The path of the config file leads into the WEB-INF folder
tomcat\webapp\myapp\WEB-INF\src\point_config.txt
Anything inside WEB-INF is protected and cannot be user-defined once the web application has launched. If you meant to read from a user-defined configuration file from the file system, please use an API like the common configuration API.
If you want to insist on keeping the file inside the WEB-INF folder, use the Class.getResourceAsStream() method to obtain the configuration instead. That would not make the configuration user-defined though.

Calling a path in java method

I have a small problem calling a path(that has the python file, that I need to run) in the following code:
Process p = Runtime.getRuntime().exec(callAndArgs,env,
new java.io.File("C:\\Users\\Balkishore\\Documents\\NetBeansProjects\\Testinstrument_Rest\\build\\web"));//excuting python file
As it can be seen from the above code, the python file is called using the path specified in java.io.file function. But it is very specific, as it can be run only in my computer. How can i make it generic, so that it is possible to run this piece of code in any computer?
Any help would be very much appreciated.

Put your python script to the location relative to your working directory and use relative path. Alternatively use configuration file or property to read the path from.

If this file is already exist in the app then you need to do
ServletContext.getRealPath("/");
which will give you the path to web root now from here you need to move relatively to reach to your file
If this is an external file
put it in ${user.home}/appname/
String filePath = System.getProperty("user.home")+File.separator+"APP_NAME"
and instruct your users to put the file in this path, or read the path from some configuration file (.properties, .conf)

Path resolution in eclipse package structure

This is a very simple question for many of you reading this, but it's quite new for me.
Here is a screenshot for my eclipse
When i run this program i get java.io.FileNotFoundException: queries.xml (The system cannot find the file specified) i tried ../../../queries.xml but that is also not working. I really don't understand when to use ../ because it means go 1 step back in dir, and in some cases it works, can anyone explain this? Also how can I refer to queries.xml here. Thanks
Note: I might even use this code on a linux box

I assume it is compiling your code into a build or classes folder, and running it from there...
Have you tried the traditional Java way for doing this:
def query = new XmlSlurper().parse( GroovySlurping.class.getResourceAsStream( '/queries.xml' ) )
Assuming the build step is copying the xml into the build folder, I believe that should work
I don't use Eclipse though, so can't be 100% sure...

Try
file = new File("src/org/ars/groovy/queries.xml");
To check the actual working directory of eclipse you can use
File f = new File(".");
System.out.println(f.getAbsolutePath());

You could try using a property file to store the path of the xml files.
This way you can place the xml files in any location, and simply change the property file.
This will not require a change/recompilation of code.
This would mean you will only need to hardcode the path of the property file.
If you prefer not hardcoding the path of the property file, then you could pass it as an argument during startup in your server setup file. (in tomcat web.xml). Every server will have an equivalent setup file where you could specify the path of the property file.
Alternatively you could specify the path of the xml in this same file, if you don't want to use property files.
This link will show you an example of reading from property files.
http://www.zparacha.com/how-to-read-properties-file-in-java/