![]() To read binary resources, you can use directly use the InputStream instance. The method returns null if the resource cannot be found or loaded. 2) For copying files in Java 6, you can either write your own code using FileChannel, FileInputStream or. Using the method Class.getResourceAsStream(String), you can get an InputStream to read the resource. Copying File from Directory to other before Java 7. Throw new Exception("resource not found: " + respath) The following code snippet shows how to load resources packed thus into the jar or war file: String respath = "/poems/Frost.txt" You can access these files and folders from your java code as shown below. Maven packs all the files and folders under main/resources into the jar file at the the root. In this article, we show you how to load these resources when the program is running. ![]() When the software is executed, it may need to load the contents of these files for some kind of processing - may be properties, sql statements, etc. These files may include configuration files, scripts and other resources needed during run time. Use FileInputStream: InputStream is new FileInputStream ('/res/example.xls') But never read from raw file input stream as this is terribly slow. ![]() When you build a java project and pack it into a jar (or a war), the files under the resources folder are included into the jar.
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