比较单元格中的2个值并更改JTable中的单元格颜色

Question

So i want to compare 2 value from 2 cells on the same row ("Initial Target" and "Result Target") from my table, and if the value is not the same the cell on "Result Target" column will turn red , but my code turn all cell to red, here's the result :

result

here's what i expect :

Expected

here's my code :

tblResult = new JTable(tableModel) {
        @Override
        public Component prepareRenderer(TableCellRenderer renderer, int row, int col) {
            Component comp = super.prepareRenderer(renderer, row, col);
            Object value1 = getModel().getValueAt(row, 2);
            Object value2 = getModel().getValueAt(row, 1);
            if (value1!=value2) {
                    comp.setBackground(Color.red);
            }
             else {
                comp.setBackground(Color.white);
            }
            return comp;
        }
    };

Answer 1

If you are doing if (value1 != value2) you are just checking whether value1 and value2 have the same reference and here they don't so, this comparison will always return true .

What you can do instead is cast these objects to String or Integer like this:

String value1 = (String) getModel().getValueAt(row, 2);
String value2 = (String) getModel().getValueAt(row, 1);

And then perform the comparison as follows:

if (!value1.equalsIgnoreCase(value2)) {}

Answer 2

You execute the logic for each columnIndex, so each rendering component has its color manipulated and is painted eg red. You should have some condition around like if(col == 2) and so only execute your color magic when the method prepareRenderer() is called for the 3rd column.

Of cause the already mentioned comparison fix if (!value1.equalsIgnoreCase(value2)) {} should be done.

By the way, you might use table.getColumnModel().getColumn(2).setCellRenderer(TableCellRenderer) to set a specific rendering logic just for this column.