I put my employeedetailxml into a folder named \"res\\raw\". When I attempt to open it by specifying the file and folder name, I get the error \"file not found\".
Let me start by saying I am fairly new to Java so forgive me if I am making obvious mistakes... I have a text file that I must read data from and split the data into separate arrays.
I am trying to get the last line of a file, but my output shows that it never finds it. I also tried looking for \"[\" which all the lines start with, but unless the jumps were perfect the program wil
I want to restore the following data from the text fi开发者_开发问答le. The problem is only one string/line I can restore, I can\'t restore the rest of the data.
When I am using FileInputStream to read an object (say a few bytes开发者_运维问答), does the underlying operation involve:
public static void writeFile(String theFileName, String theFilePath) 开发者_StackOverflow{ try { File currentFile = new File(\"plugins/mcMMO/Resources/\"+theFilePath+theFileName);
Consider these two functions: Function A takes inputStream as parameter. public void processStream(InputStream stream)
I am developing an encryption tool, and for our encrypted file format I am using Base64 to encode data. I am using apache commons codec to decode files using a Base64InputStream wrapped around a FileI
I am an amateur programmer developing for android. I am just trying get the basics down right now, but I am having an error and I don\'t know why.
I am in the process of writing an application that processes a huge number of integers from a binary file (up to 50 meg). I need to do it as quickly as possible and the main performance issue is the d
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