SQL查询以统计在日期范围内与某些条件匹配的响应数，并按每天分组显示

Question

I have the following set of survey responses in a table.

It's not very clear but the numbers represent the 'satisfaction' level where:

0 = happy
1 = neutral
2 = sad

+----------+--------+-------+------+-----------+-------------------------+
| friendly | polite | clean | rate | recommend |      booking_date       |
+----------+--------+-------+------+-----------+-------------------------+
|        2 |      2 |     2 |    0 |         0 | 2014-02-03 00:00:00.000 |
|        1 |      2 |     0 |    0 |         2 | 2014-02-04 00:00:00.000 |
|        0 |      0 |     0 |    1 |         0 | 2014-02-04 00:00:00.000 |
|        1 |      1 |     2 |    0 |         2 | 2014-02-04 00:00:00.000 |
|        0 |      0 |     1 |    2 |         1 | 2014-02-04 00:00:00.000 |
|        2 |      2 |     0 |    2 |         0 | 2014-02-05 00:00:00.000 |
|        2 |      1 |     1 |    0 |         2 | 2014-02-05 00:00:00.000 |
|        1 |      0 |     1 |    2 |         0 | 2014-02-05 00:00:00.000 |
|        0 |      1 |     1 |    1 |         1 | 2014-02-05 00:00:00.000 |
|        1 |      0 |     2 |    2 |         0 | 2014-02-05 00:00:00.000 |
+----------+--------+-------+------+-----------+-------------------------+

For each day I need the totals of each of the columns matching each response option. This will answer the question: "How may people answered happy, neutral or sad for each of the available question options".

I would then require a recordset returned such as:

+------------+----------+------------+--------+----------+------------+--------+
|    Date    | FriHappy | FriNeutral | FriSad | PolHappy | PolNeutral | PolSad |
+------------+----------+------------+--------+----------+------------+--------+
| 2014-02-03 |        0 |          0 |      1 |        0 |          0 |      1 |
| 2014-02-04 |        2 |          2 |      0 |        2 |          1 |      1 |
| 2014-02-05 |        1 |          2 |      2 |        2 |          2 |      1 |
+------------+----------+------------+--------+----------+------------+--------+

This shows that on the 4th two responders answered "happy" for the "Polite?" question, one answered "Neutral" and one answered "sad".

On the 5th, one responder answered "happy" for the Friendly option, two choose "neutral" and two chose "sad".

I really wish to avoid doing this in code but my SQL isn't great. I did have a look around but couldn't find anything matching this specific requirement.

Obviously this is never going to work (nice if it did) but this may help explain:

SELECT cast(booking_date as date) [booking_date], 
COUNT(friendly=0) [FriHappy],
COUNT(friendly=1) [FriNeutral],
COUNT(friendly=2) [FriSad]
FROM [u-rate-gatwick-qsm].[dbo].[Questions]
WHERE booking_date >= '2014-02-01'
AND booking_date <= '2014-03-01'
GROUP BY cast(booking_date as date) 

Any pointers would be much appreciated.

Many thanks.

Answer 1

Here is a working version of your sample query:

SELECT cast(booking_date as date) as [booking_date], 
       sum(case when friendly = 0 then 1 else 0 end) as [FriHappy],
       sum(case when friendly = 1 then 1 else 0 end) as [FriNeutral],
       sum(case when friendly = 2 then 1 else 0 end) as [FriSad]
FROM [u-rate-gatwick-qsm].[dbo].[Questions]
WHERE booking_date >= '2014-02-01' AND booking_date <= '2014-03-01'
GROUP BY cast(booking_date as date) 
ORDER BY min(booking_date);

Your expression count(friendly = 0) doesn't work in SQL Server. Even if it did, it would be the same as count(friendly) -- that is, the number of non- NULL values in the column. Remember what count() does. It counts the number of non-NULL values.

The above logic says: add 1 when there is a match to the appropriate friendly value.

By the way, SQL Server doesn't guarantee the ordering of results from an aggregation, so I also added an order by clause. The min(booking_date) is just an easy way of ordering by the date.

And, I didn't make the change, but I think the second condition in the where should be < rather than <= so you don't include bookings on March 1st (even one at exactly midnight).