How should I load files into my Java application

Question

User · Answer

What are you loading the files for - configuration or data  like an input file  or as a resource    If as a resource  follow the suggestion and example given by Will and Justin  If configuration  then you can use a ResourceBundle or Spring  if your configuration is more complex   If you need to read a file in order to process the data inside  this code snippet may help BufferedReader file   new BufferedReader new FileReader filename   and then read each line of the file using file readLine     Don t forget to close the file

User · Answer

The short answer  Use one of these two methods    Class getResource String  Class getResourceAsStream String    For example   InputStream inputStream   YourClass class getResourceAsStream  image jpg      --  The long answer  Typically  one would not want to load files using absolute paths  For example  don   t do this if you can help it   File file   new File  C   Users  Joe  image jpg      This technique is not recommended for at least two reasons  First  it creates a dependency on a particular operating system  which prevents the application from easily moving to another operating system  One of Java   s main benefits is the ability to run the same bytecode on many different platforms  Using an absolute path like this makes the code much less portable   Second  depending on the relative location of the file  this technique might create an external dependency and limit the application   s mobility  If the file exists outside the application   s current directory  this creates an external dependency and one would have to be aware of the dependency in order to move the application to another machine  error prone    Instead  use the getResource   methods in the Class class  This makes the application much more portable  It  can be moved to different platforms  machines  or directories and still function correctly

User · Answer

getResource is fine  but using relative paths will work just as well too  as long as you can control where your working directory is  which you usually can    Furthermore the platform dependence regarding the separator character can be gotten around using File separator  File separatorChar  or System getProperty  file separator

User · Answer

public static String loadTextFile File f        try           BufferedReader r   new BufferedReader new FileReader f            StringWriter w   new StringWriter             try               String line   reader readLine                while  null    line                    w append line  append   n                    line   r readLine                               return w toString              finally               r close                w close                    catch  Exception ex            ex printStackTrace             return

User · Answer

I haven t had a problem just using Unix-style path separators  even on Windows  though it is good practice to check File separatorChar   The technique of using ClassLoader getResource   is best for read-only resources that are going to be loaded from JAR files  Sometimes  you can programmatically determine the application directory  which is useful for admin-configurable files or server applications   Of course  user-editable files should be stored somewhere in the System getProperty  quot user home quot   directory

User · Answer

public byte   loadBinaryFile  String name        try            DataInputStream dis   new DataInputStream new FileInputStream name            byte   theBytes   new byte dis available             dis read theBytes  0  dis available             dis close            return theBytes        catch  IOException ex              return null

[java] How should I load files into my Java application?

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