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

Question

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)

thanx, but how should I wrote its path in reportSource field? — zigzag
– zigzag, Commented Oct 20, 2015 at 19:42
Google how to use jar resources. You can read it as a stream from your jar — user1231232141214124
– user1231232141214124, Commented Oct 20, 2015 at 19:42

AbtPst · Accepted Answer · 2015-10-20 19:43:59Z

1

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

answered Oct 20, 2015 at 19:43

AbtPst

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nabeel · Accepted Answer · 2015-10-20 20:51:42Z

0

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

answered Oct 20, 2015 at 20:51

nabeel

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