Cannot convert Array to List of String correctly

Question

There is string with fixed length and it has to be spitted by 15 characters each record. The results should be placed in List, however it seems like whole string always placed in 0 position in List.

Arrays.asList(a1.substring(1,324).split("[a-zA-Z]{20}"))

Why is that?

UPDATE:

List<String> l =  Arrays.asList("1111111111     1119999999                                                                                              ".split("[0-9]{15}"));

I am not saying you should change it, I am asking what do you think it does (what results do you think you should get)? Also which part it its documentation makes you think that such results should be expected? — Pshemo
– Pshemo, Commented Sep 9, 2014 at 21:00
@Pshemo, it splits string if it finds consecutive 20 characters in set [a-zA-Z]. White spaces won't match. Right? — J.Olufsen
– J.Olufsen, Commented Sep 9, 2014 at 21:02

user3487063 · Accepted Answer · 2014-09-09 21:10:36Z

3

split regex should be :

String[] arr = str.split("(?<=\\G.{20})");

The above split str for every 20 chars

For example below code splits str for every 15 chars:

 String str ="hkdhadhkshdkhskhdkashdkasgi2oyeihsadkahdkashdlkhas";
             List<String> list = Arrays.asList(str.split("(?<=\\G.{15})"));
            System.out.println(list);

prints:

[hkdhadhkshdkhsk, hdkashdkasgi2oy, eihsadkahdkashd, lkhas]

edited Sep 9, 2014 at 21:10

answered Sep 9, 2014 at 21:03

user3487063

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6 Comments

J.Olufsen Over a year ago

Nice and succinct! I am not really understand this part ?<=\\G.

Bohemian Over a year ago

+1 Nice split regex. There are not many who know this needs \G, or even what \G does.

Pshemo Over a year ago

@RCola \G represents end of previous match, or if there is no previous match (in case where we are looking for first match) it represents start of the String (is equivalent of ^).

Bohemian Over a year ago

@pshemo this reminds me of your stellar answer, which was how I learned about \G. It is still my favourite answer on the site - it changed my life :)

user3487063 Over a year ago

In other words it means , match an empty string from the end of previous match after 15 characters.

|

jdphenix · Accepted Answer · 2014-09-09 21:06:06Z

From your question:

There is string with fixed length and it has to be spitted by 15 characters each record. The results should be placed in List.

String.split() isn't really suitable for something like this. It's designed to split on a delimiter, which "every 15 characters" is not.

You're looking for the functionality in String.substring(), which returns a String that is the sequence between [beginIndex, endIndex).

With this method:

public static List<String> splitByIndex(String toSplit, int index) { 
    List<String> result = new ArrayList<String>(); 
    for (int i = 0; i < toSplit.length(); i += index) { 
        if (i + index < toSplit.length()) {
            result.add(toSplit.substring(i, i + index));
        } else {
            result.add(toSplit.substring(i, toSplit.length() - 1));
        }
    }
    return result; 
}

You can split a String into a List<String> by a given number of characters. This example code:

String a1 = "I'm the fixed length string that needs to be split by 15 characters.";
List<String> list = splitByIndex(a1, 15);
System.out.println(list);

will output:

[I'm the fixed l, ength string th, at needs to be , split by 15 cha, racters]

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