Want to win a PS4? Go Premium and enter to win our High-Tech Treats giveaway. Enter to Win

x
?
Solved

Splitting a file into batches

Posted on 2012-03-15
6
Medium Priority
?
276 Views
Last Modified: 2012-03-16
Hi:
I have a text file, extracted from a database. It is a set of instructions for a saw.
Note that column eight, beginning 169... changes from 169022 to 169023, etc.
Is there any way of breaking the file at each change of batch; creating a new file, and each file having the name of the  batch it contains. So currently the file is called 169022. I need that to be JUST 169022, with different files for each subsequent batch.
I can from the database create a text file with a list of the batches if needs be.
Java is good - but will accept anything (c++, c#, vb, etc!)

Thanks!
0
Comment
Question by:ClaytonGlass
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 3
  • 2
6 Comments
 

Author Comment

by:ClaytonGlass
ID: 37724384
Sorry...forgot the file...
0
 

Author Comment

by:ClaytonGlass
ID: 37724396
Sorry...forgot the file...
169022.txt
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 37725647
You might try this class, which contains a rather crude implementation of T for your file (but it seems to work)
import java.io.*;

import java.util.*;


public class ComparableSublistRetriever<T extends Comparable<T>>
    implements Iterator<List<T>> {
    private List<T> delegate;
    private T previous;
    private T current;
    private int index;

    public ComparableSublistRetriever(List<T> items) {
        delegate = new ArrayList<T>(items);
        Collections.sort(delegate);
    }

    @Override
    public boolean hasNext() {
        return index < delegate.size();
    }

    @Override
    public List<T> next() {
        List<T> sublist = new ArrayList<T>();
        T current = delegate.get(index++);
        sublist.add(current);
	previous = current;
	while(index < delegate.size() && (current = delegate.get(index)).equals(previous)) {
            sublist.add(current);
            previous = current;
	    index++;

        }
        return sublist;
    }

    @Override
    public void remove() {
        ;
    }

    static class CsvLine implements Comparable<CsvLine> {
	private String[] array;

	public CsvLine(String[] array) {
	    this.array = array;
	}

	public int compareTo(CsvLine other) {
	    return array[7].compareTo(other.array[7]);
	}

	public boolean equals(Object other) {
	    return array[7].equals(((CsvLine)other).array[7]);
	}

	public String toString() {
	    return Arrays.toString(array);
	}
    }

    public static void main(String[] args) throws Exception {
	List<CsvLine> csvLines = new ArrayList<CsvLine>();
	Scanner s = new Scanner(new File(args[0]));
	while(s.hasNextLine()) {
	    csvLines.add(new CsvLine(s.nextLine().split("\\s+", 9)));
	}
	s.close();
        ComparableSublistRetriever<CsvLine> r = new ComparableSublistRetriever<CsvLine>(csvLines);

        while (r.hasNext()) {
            System.out.println(r.next());
        }
    }
}

Open in new window

0
Concerto's Cloud Advisory Services

Want to avoid the missteps to gaining all the benefits of the cloud? Learn more about the different assessment options from our Cloud Advisory team.

 
LVL 47

Expert Comment

by:for_yan
ID: 37726247
This is the code for the complete program which does it;
See attached the files produeced from your input file
import java.io.BufferedReader;
import java.io.FileOutputStream;
import java.io.FileReader;
import java.io.PrintStream;
import java.util.ArrayList;
import java.util.HashMap;

public class SplitFile {
    public SplitFile() {
        try{
            HashMap <String, ArrayList<String>> map = new HashMap<String, ArrayList<String>>();
            ArrayList<String> names = new ArrayList<String>();

            BufferedReader br = new BufferedReader((new FileReader("169022.txt")));

            String buff = "";

            while((buff=br.readLine()) != null){
                String [] items = buff.split("[\\s]+");
                if(items.length < 3)continue;



                 if(map.get(items[7]) != null){
                     ArrayList<String> ar = map.get(items[7]);
                     ar.add(buff);
                       map.put(items[7], ar);

                 }      else
                 {
                     ArrayList<String> ar = new ArrayList<String>();
                     ar.add(buff);
                     System.out.println(items[7]);
                     names.add(items[7]);
                       map.put(items[7], ar);
                 }

                


            }

            br.close();

            for(int j=0; j< names.size(); j++){
                String name = names.get(j);
                PrintStream ps = new PrintStream(new FileOutputStream(name + "_out.txt"));

                ArrayList<String> ar = map.get(name);
                for(String s: ar){
                    ps.println(s);
                }

                 ps.close();
            }






        }  catch(Exception ex){
            ex.printStackTrace();

        }


    }

    public static void main(String[] args) {
        new SplitFile();
        }
}

Open in new window

169022-out.txt
169023-out.txt
169024-out.txt
169025-out.txt
169026-out.txt
169027-out.txt
0
 
LVL 47

Accepted Solution

by:
for_yan earned 1200 total points
ID: 37726390
This is even simpler varian which does not depend on the size fo the file and can handle files of any size.

Output files are attached.

import java.io.*;
import java.util.ArrayList;
import java.util.HashMap;

public class SplitFile1 {

       public SplitFile1() {
        try{

            String sep = System.getProperty("line.separator");

            BufferedReader br = new BufferedReader((new FileReader("169022.txt")));
            String curFile = null;
            FileWriter fw = null;

            String buff = "";

            while((buff=br.readLine()) != null){
                String [] items = buff.split("[\\s]+");
                if(items.length < 3)continue;
                String name = items[7];
                if(!name.equals(curFile)){

                    curFile = name;
                    if(fw != null)fw.close();
                    fw = new FileWriter(new File(curFile+ "_out1.txt"), true);
                    

                }
               if(fw != null) fw.write(buff + sep);


            }

            br.close();
            if(fw != null) fw.close();

        } catch(Exception ex){
            ex.printStackTrace();
        }




}

    public static void main(String[] args) {
        new SplitFile1();
    }


}

Open in new window

169022-out1.txt
169023-out1.txt
169024-out1.txt
169025-out1.txt
169026-out1.txt
169027-out1.txt
0
 

Author Closing Comment

by:ClaytonGlass
ID: 37728308
Thank You Very Much!
0

Featured Post

Free Tool: Path Explorer

An intuitive utility to help find the CSS path to UI elements on a webpage. These paths are used frequently in a variety of front-end development and QA automation tasks.

One of a set of tools we're offering as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Introduction This article is the second of three articles that explain why and how the Experts Exchange QA Team does test automation for our web site. This article covers the basic installation and configuration of the test automation tools used by…
The article shows the basic steps of integrating an HTML theme template into an ASP.NET MVC project
This theoretical tutorial explains exceptions, reasons for exceptions, different categories of exception and exception hierarchy.
This tutorial will introduce the viewer to VisualVM for the Java platform application. This video explains an example program and covers the Overview, Monitor, and Heap Dump tabs.
Suggested Courses

604 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question