Reading and Writing in the middle of a file is as simple as using a RandomAccessFile in Java. RandomAccessFile, despite its name, is more like an InputStream and OutputStream and less like a File. It allows you to read or seek through bytes in a file and then begin writing over whichever bytes you care … Read more
The underlying data source for some implementations of InputStream can signal that the end of the stream has been reached, and no more data will be sent. Until this signal is received, read operations on such a stream can block. For example, an InputStream from a Socket socket will block, rather than returning EOF, until … Read more
I’ve searched all over the web and after reading lot of docs regarding connection timeout exception, the thing I understood is that, preventing SocketTimeoutException is beyond our limit. One way to effectively handle it is to define a connection timeout and later handle it by using a try-catch block. Hope this will help anyone in … Read more
When possible, you should not store the entire contents of a file to be served in memory. Instead, aquire an InputStream for the data, and copy the data to the Servlet OutputStream in pieces. For example: ServletOutputStream out = response.getOutputStream(); InputStream in = [ code to get source input stream ]; String mimeType = [ … Read more
When closing chained streams, you only need to close the outermost stream. Any errors will be propagated up the chain and be caught. Refer to Java I/O Streams for details. To address the issue However, if flush() throws a runtime exception for some reason, then out.close() will never be called. This isn’t right. After you … Read more
I assume you want to be able to use step-through debugging from Eclipse. You can just run the classes externally by setting the built classes in the bin directories on the JRE classpath. java -cp workspace\p1\bin;workspace\p2\bin foo.Main You can debug using the remote debugger and taking advantage of the class files built in your project. … Read more
CodeSource src = MyClass.class.getProtectionDomain().getCodeSource(); if (src != null) { URL jar = src.getLocation(); ZipInputStream zip = new ZipInputStream(jar.openStream()); while(true) { ZipEntry e = zip.getNextEntry(); if (e == null) break; String name = e.getName(); if (name.startsWith(“path/to/your/dir/”)) { /* Do something with this entry. */ … } } } else { /* Fail… */ } Note that … Read more
Place your text file in the /assets directory under the Android project. Use AssetManager class to access it. AssetManager am = context.getAssets(); InputStream is = am.open(“test.txt”); Or you can also put the file in the /res/raw directory, where the file will be indexed and is accessible by an id in the R file: InputStream is … Read more
Just implement Serializable If you’re getting a NotSerializableException like follows, java.io.NotSerializableException: bean.ProjectAreaBean then it simply means that the class as identified by the fully qualified name in the exception message (which is bean.ProjectAreaBean in your case) does not implement the Serializable interface while it is been expected by the code behind. Fixing it is relatively … Read more
Code : public class JavaApplication { public static void main(String[] args) { System.out.println(“Working Directory = ” + System.getProperty(“user.dir”)); } } This will print the absolute path of the current directory from where your application was initialized. Explanation: From the documentation: java.io package resolve relative pathnames using current user directory. The current directory is represented as … Read more