Array.toString data handling in Java

Question

I'm confused the way data has been manipulated here in this case, thought will seek someones advice.

arrayDataOne = [ HEllO, NAME, ADDRESS, ZIPCODE ] 

i.e  
[0] = "HELLO"   
[1] = "NAME"   
[2] = "ADDRESS"   
[3] = "ZIPCODE"

String arrayInString = ( Arrays.toString(arrayDataOne));

i.e
[ HEllO, NAME, ADDRESS, ZIPCODE ] 

String[] arraySplit = arrayInString.split("\\|");

i.e
[0] = [[HEllO, NAME, ADDRESS, ZIPCODE ] ]

Question,
Why does string value displays as array though logically it's not stored as array in arrayInString ?

Why does arraySplit holds data with in [[ ****data**** ]] ? seems like array of array?

Edited: ( added more answer details )

To handle string without "[" "]" : read link

Luiggi Mendoza · Accepted Answer · 2013-04-29 00:02:14Z

4

Because that's what .toString does. Read the doc http://docs.oracle.com/javase/6/docs/api/java/util/Arrays.html

Returns a string representation of the contents of the specified array. The string representation consists of a list of the array's elements, enclosed in square brackets ("[]"). Adjacent elements are separated by the characters ", " (a comma followed by a space). Elements are converted to strings as by String.valueOf(double). Returns "null" if a is null.
Because it does not match anything when you try to split it. The result therefore is a single bucket the string itself

edited Apr 29, 2013 at 0:02

Luiggi Mendoza

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answered Apr 29, 2013 at 0:01

marcadian

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3 Comments

Mad-D Over a year ago

so if i'm interested in just plain string of HEllO, NAME, ADDRESS, ZIPCODE without [ ], i should do regex replace on it ? or is it a bad practice ?

marcadian Over a year ago

If you want it as array, then do not use .toString. What actually are you trying to do?

Mad-D Over a year ago

thanks i did find the answer, used StringBuilder to create string.

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