can't pass fourth task requirement

Question about the task

Two in one

Java Core, Level 8, Lesson 11

Resolved

any hint for appending content to a file without erasing previous file content? thank you.

Read 3 file names from the console.
Write the contents of the second file to the first file, and then append the contents of the third file to the first file.
Close the streams.

Requirements:

The program should read a file name three times from the console.
Create an output stream for the first file. For the other two, create input streams.
The contents of the second file must be copied to the first file.
The contents of the third file must be appended to the first file (after the second file has been written to the first file).
The file streams must be closed.

package com.codegym.task.task18.task1818; /* Two in one */ import java.io.*; public class Solution { public static void main(String[] args) { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); try { String file1 = br.readLine(); String file2 = br.readLine(); String file3 = br.readLine(); FileOutputStream outputStream = new FileOutputStream(file1); FileInputStream inputStream1 = new FileInputStream(file2); FileInputStream inputStream2 = new FileInputStream(file3); byte[] buffer = new byte[1000]; while(inputStream1.available() > 0) { int i = inputStream1.read(buffer); outputStream.write(buffer); } byte[] buffer1 = new byte[1000]; while(inputStream2.available() > 0) { int i = inputStream2.read(buffer); } byte[] buffer2 = new byte[buffer.length + buffer1.length]; int index = buffer.length; for (int i = 0; i < buffer.length; i++) { buffer2[i] = buffer[i]; } for (int i = 0; i < buffer1.length; i++) { buffer2[i + index] = buffer1[i]; } for(int b : buffer2){ outputStream.write(b); } outputStream.close(); inputStream1.close(); inputStream2.close(); } catch(IOException e){} } }

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Thang Za Thang

Level 18 , Melbourne, Australia

5 August 2020, 02:45

This worked for me! BufferedReader reader = new BufferedReader(new InputStreamReader(System.in)); String file1 = reader.readLine(); String file2 = reader.readLine(); String file3 = reader.readLine(); reader.close(); try (FileOutputStream outputStream = new FileOutputStream(file1); FileInputStream inputStream1 = new FileInputStream(file2); FileInputStream inputStream2 = new FileInputStream(file3); ){ byte[] buffer = new byte[inputStream1.available()]; int count = inputStream1.read(buffer); outputStream.write(buffer, 0, count); // Contents of the file2 is written to the first file, file1. //buffer buffer = new byte[inputStream2.available()]; count = inputStream2.read(buffer); outputStream.write(buffer, 0, count); // Contents of the file3 is written to the first file, file1. outputStream.close(); inputStream1.close(); inputStream2.close(); }

Guadalupe Gagnon

Level 37 , Tampa, United States

7 April 2019, 23:29

First problem that I see is that your buffers are only 1000 bytes long. The files may be larger than that, and when you read 1000 bytes at a time, only the last .read() will be in those arrays. When you combine them into buffer2, you may be losing data. Use this:

byte[] buffer = new byte[FileInputStream.available()];

The .available() method return an int that is the total bytes of the file. In this task you do not need to worry about 'appending'. When an output stream is created it will create a file that is input, or clear the file if one exists already. Because this only happens during creation you can write 'buffer' and then 'buffer1' to the file without data being lost. If the task had you append file 3 to file 2, then you would have to figure out how not to erase data (you need to set a flag in the constructor). Make sure to close the buffered reader after reading 3 file names. After that, your solve looks like this: 1. create inStr1 2. create byte array, make its size inStr1.available() 3. read everything into buffer, you DO NOT need a loop as the buffer is already the exact size you need 4. close inStr1 5. create inStr2 6-8 copy everything from 2-4 7. create outStr 8. write buffer1 - once again you do not need to loop, just outStr.write(buffer) will write it all 9. repeater for buffer2 10. close outStr. Get in the habit of only having 1 stream open at a time (when possible). Streams are very resource intensive and, while not going to affect these small task, can cause major problems in large complex programs.

Guadalupe Gagnon

Level 37 , Tampa, United States

7 April 2019, 23:40

Example:

byte[] buffer = new byte[inputStream1.available()];

//while(inputStream1.available() > 0) {
/*int i = */
inputStream1.read(buffer);
outputStream.write(buffer);
//}

This is all your code, modified, and valid. You should be able to use it to figure out how to do everything I said above.

Djole

Level 41 , Belgrade, Serbia

8 April 2019, 09:26

Thanks a lot! That was very informative.