Getting the Difference in JTables my algo is messed

Question

Im trying to get the difference of two JTables...in other words if Table B has the rowdata of Table A, then Table C should not have it...Table C should have all the data in Table A - the data in Table B
heres my code:
//in a subclass of JTable
public Object[][] getDifferenceData(JTable table){
    DefaultTableModel thisModel = (DefaultTableModel)this.getModel();
    DefaultTableModel userModel = (DefaultTableModel)table.getModel();
    int differenceRows = thisModel.getRowCount() - userModel.getRowCount();
    Object[][] newData = new Object[differenceRows][2];
    Object symbol = new Object();
    Object description = new Object();
     for (int i=0; i<differenceRows; i++){
        for (int j=0; j<userModel.getRowCount()-1; j++){
            if ((symbol = thisModel.getValueAt(i,0)) != userModel.getValueAt(j,0)){
                //symbol in portfolio model != symbol in user portfolio
                //so new difference table model should have it
                description = thisModel.getValueAt(i,1);
                newData[0] = symbol;
newData[i][1] = description;
return newData;
heres what im trying to do
1) new Data[][] is set to size bigger table row count - smaller tables
for all new row size
2)compare table A data in the current row and first column to all the rows/first column of table B
3) if the data is NOT the same, the copy that row data from Table A into a new 2D data vector to the appropriate row
4) return the new data[][] for new jtable creation
but what happens is that the difference table (C) contains the corrent number of rows...but it still contains the data found in table B, and the new table C is just table C with the right number of rows cut off
my algo is severly messed :)
any help would be appreciated..thanks

Answer 1

hey klepto,
your not using .equals() on the symbol Object, this means that all it is doing is seeing if they refer to the same instance of the object, not if the to instances have the same value.
The other problem is that you presume that the difference of the two collections is based on their respective size. What if they contain totaly different values? Then the total size would be the sum of both collections.
Lets presume that the symbol is a String. Then I would do the following:
//in a subclass of JTable
public Object[][] getDifferenceData(JTable table, DefaultTableModel thisModel, DefaultTableModel userModel){
    Object[][] newData = null;
    String symbolA = null;
    String symbolB = null;
    String description = null;
    SymbolData symbolData = null;
    Vector newList = new Vector();
    int nbrRowsA = thisModel.getRowCount();
    int nbrRowsB = userModel.getRowCount();
    for(int i = 0; i < nbrRowsA; i++) {
        symbolA = (String)thisModel.getValue(i, 0);
        for(int j = 0; j < userModel.getRowCount() - 1; j++) {
            symbolB = (string)userModel.getValue(i, 0);
            if(!symbolA.equals(symbolB)){
                description = (String)thisModel.getValueAt(i,1);
                newList.add(new SymbolData(symbol, description));
    if(newList.size() > 0) {
        int size = newList.size();
        newData = new String[size][2];
        for(int i = 0; i < size; i++) {
            symbolData  = (SymbolData)newList.get(i);
            newData[0] = symbolData.getSymbol();
newData[i][1] = symbolData.getDescription();
return newData;
class SymbolData {
String symbol;
String description;
MyDataClass(String symbol, String description) {
this.symbol = symbol;
this.description = description;
String getSymbol() {
return symbol;
String getDescription() {
return description;
String[] getData() {
String[] data = new String[2];
data[0] = symbol;
data[1] = description;
return data;
Hmm, that should do the job. Not the fastest code, I'm sure there
could be some improvements with the array handling. But it should
get you started.
Also I have not attempted to compile the above so I suspect you
will find problems (only on a good day do I get zero compile errors
when I first write something, good days are rare :-) ).
James.