Why i get Null values in my second counter using case statement - mysql

Question

The first case statement i got the correct result but in the second one
Why i got an NULL result Where my second case statement the counter = 2
this is the result i have an image
Query Result that i got Null data in second statement when i grouped by on my date
SELECT DISTINCT date,log,
CASE
WHEN note = 'HOLIDAY' AND counter = 1
THEN 'HOLIDAY'
END note1,
CASE
WHEN note = 'HOLIDAY' AND counter = 2
THEN 'HOLIDAY'
END note2,
FROM timesheet
WHERE timesheet.empid='40' AND date <= CURDATE() AND YEAR(date)= YEAR(CURDATE())
AND MONTH(date) = MONTH(CURDATE())
GROUP BY date
ORDER BY date DESC;

Answer 1

You're using GROUP BY wrong. The rule is that each column in your SELECT clause is either also in your GROUP BY clause or an aggregate function (like count, min, max, avg) must be applied to it.
When you don't follow this rule, a random row for each group is displayed. In your case, when you really have data with note = 'HOLIDAY' AND counter = 2, the rows for the group might look like this
NULL
HOLIDAY
NULL
NULL
but after collapsing (when it's outputted by the select), just the first row is displayed, therefore the NULL value.
Try it like this:
SELECT date,
MIN(log), /*or maybe you want to group by this column, too? */
MAX(CASE
WHEN note = 'HOLIDAY' AND counter = 1
THEN 'HOLIDAY'
END) note1,
MAX(CASE
WHEN note = 'HOLIDAY' AND counter = 2
THEN 'HOLIDAY'
END) note2,
FROM timesheet
WHERE timesheet.empid='40' AND date <= CURDATE() AND YEAR(date)= YEAR(CURDATE())
AND MONTH(date) = MONTH(CURDATE())
GROUP BY date
ORDER BY date DESC;
Also note, that I removed the DISTINCT. Your GROUP BY already does that.