Hi rw,
Conditional formatting rules can be set only for individual cells, not for whole rows. The rules compare the value in the cell with a fixed value, written into the rule, or with the value in another cell.
In each row of your table, you have five cells, three of which are checkboxes, limited to containing one of two values, 'true' or 'false. /With only two possible values, I see no possible set of conditions that would provide four distinct comparisons for the 'true' or 'false' state of these cells.
The fallback here is to use two tables, one behind the other, and to highlight the single cell rows in the back table. Here's an example, using the checkbox values on your table:
The original table is on the left, the 'high;ights' table is on the right.
I've set the default fill on all rows of the highlight table to the shade of blue you specified for 'none checked'.
The formula below the tables is entered in row 2 of the highlights table, then filled down to row 10. Each IF part returns the larger number in its part (100, 10, or 1) if its checkbox is checked, or zero if its checkbox is unchecked. SUM then sums the three numbers and returns a sum with 1 indicating a checked cell and 0s indicating an unchecked cell. (Leading zeros are not displayed.)
Each of the numbers (except 0) for one of the specified patterns is set as the condition for one of the conditional highlighting rules applied to the rows of the highlighting table. Zero (or any pattern creating a sum for which there is no rule set) will leave the cell fill at its default 'blue fill'.
Here are the rules applied to the table as it stands above:
Note that Rule 1 applies the same colour to both the cell fill and the text in the cell, hiding the 10 calculated by the formula. Before placing this table behind the one containing the checkboxes, the custom style for each of the other rules will also include matching colour for fill and text, as will the default format for all of the rows.
Last step is to move the highlighting tale to the back, then slide it behind the original table. (I also checked a second set of rows, using the same patterns)
Regards,
Barry
