Error Parsing a Csv File with Java

Question

I have a csv file with values like these:

1/1/1983;1,7;-3;8;-0,7;84;4;2;11;0;1030;0;0
2/1/1983;2,7;-2;8,4;1,9;94;2;2;15;0;1027;0;0
3/1/1983;4,1;-0,4;11,3;3,1;93;3;3;13;0;1030;0;0
4/1/1983;7,6;1,3;15;5,1;84;9;8;28;0;1027;0;0
5/1/1983;5,6;1,4;10;5,1;97;2;2;11;0;1023;0;0
6/1/1983;7;5,5;7,5;7;100;1;3;9;0;1028;0;0
7/1/1983;7,7;5;13,4;7,1;96;1;4;20;0;1029;0;0
8/1/1983;7,9;7;15,5;7,4;97;2;7;24;0;1029;0;1
9/1/1983;6,7;1;10,3;4,1;83;8;15;44;0;1033;0;0,3
10/1/1983;2,2;-1,9;8;0,4;88;8;4;13;0;1036;0;0
11/1/1983;0,7;-3,4;6,4;-1,2;87;3;1;13;0;1038;0;0
12/1/1983;0,2;-4,7;8;-1,7;87;6;4;9;0;1037;0;0
13/1/1983;1,7;-5,2;11,1;-0,1;88;4;3;15;0;1032;0;

So i have found on a website a Csv Parser implementation:

import java.io.File;
import java.io.FileNotFoundException;
import java.util.ArrayList;
import java.util.Scanner;

public class ScannerExample 
{
    public static void main(String[] args) throws FileNotFoundException 
    {
        //Array
        ArrayList<String> weather = new ArrayList<String>();
        //Get scanner instance
        Scanner scanner = new Scanner(new File("C:/csv/meteo2.csv"));

        //Set the delimiter used in file
        scanner.useDelimiter(";");

        //Get all tokens and store them in some data structure
        //I am just printing them
        while (scanner.hasNext()) 
        {
            weather.add(scanner.next());
        }
         System.out.println(weather.get(12));
        //Do not forget to close the scanner  
        scanner.close();
    }
}

But I Have a problem with the last element of one line and the first element of the successive line : Infact when in the code i try to print the twelfth element (that must be 0). It prints

0
2/1/1983

But it's considered as only one element. There is a solution to that?

You can read the file by lines (scanner.nextLine()) and then split the induvidual lines. — Natix
– Natix, Commented Sep 24, 2013 at 10:33

jlordo · Accepted Answer · 2013-09-24 10:33:44Z

4

If you want the semicolon and the linebreak as a delimiter, you should use

scanner.useDelimiter("[;\n]");

as Scanner#useDelimiter(String pattern) expects a regular expression as the parameter.

answered Sep 24, 2013 at 10:33

jlordo

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4 Comments

alessandrob Over a year ago

sorry, it works but after the last element (the 0) remain a space like above, altough now the subsequent date is correctly another element of the list. How can i solve?

jlordo Over a year ago

@Markviduka: Sorry, I don't understand what you mean.

alessandrob Over a year ago

Above i have this string "0 (newline) 2/1/1983". With your code i solved the problem, and now, "2/1/1983" it's a new string. However, the "0" string have the same problem of above and it's "0 (newline) " . Hope you understand

jlordo Over a year ago

@Markviduka: No, when \n is one of the delimiters, it won't be part of a token. You probably insert the newline character when printing...

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