I have following
var arrayT = new Array();
arrayT['one'] = arrayT['two'] = new Array();
arrayT['one']['a'] = arrayT['one']['b'] = '';
arrayT['two'] = arrayT['one'];
arrayT['two']['a'] = 'test';
console.log(arrayT);
In console I have
[one][a]='test'
[one][b]=''
[two][a]='test'
[two][b]=''
Why?
The line
arrayT['one'] = arrayT['two'] = new Array();
creates a single shared array object. Each "inner" array in your two-dimensional array is really just a reference to the same object, so altering one "inner" array will necessarily affect the other in the exact same way.
Instead, create two separate arrays:
arrayT['one'] = new Array();
arrayT['two'] = new Array();
Futhermore, even if you implement that change, the line:
arrayT['two'] = arrayT['one'];
will create the same problem -- arrayT['two'] and arrayT['one'] will point to the same object, possibly causing future problems of a similar nature (e.g., altering arrayT['two']['a'] on the next line will alter arrayT['one']['a'], since they point to the same object).
arrayT['one'] and arrayT['two'] are the same array, any change you make to one will affect the other.
To fix this, create separate arrays:
arrayT['one'] = new Array();
arrayT['two'] = new Array();
This issue of multiple references to the same array happens when you use arrayT['one'] = arrayT['two'] = new Array(), but arrayT['two'] = arrayT['one'] will create the same issue. To make arrayT['two'] a copy of arrayT['one'] you should use the following:
arrayT['two'] = arrayT['one'].slice();
While apsillers is right with his answer, a more accurate replacement would be to simply write arrayT['two'] = arrayT['one'].slice(0)
leaving out the assignment to arrayT['two'] in line 2.
Note, that this is only a shallow copy. Depending on your needs you might need to do a deep copy, if you should decide to use mutable objects later.
