SQL和分组依据，选择问题

Question

Hello and sorry for uninformative title. Could not think of better. So, my problem is thus. I have two key variables in two tables, table1 has only key1 and table2 has both key1 and key2 to link key1 to key2. Table1 also has a year-specific variable.

Now here is the problem. What I want to know is the list of all key1 variables that are linked to such key2 variables that have more than one key1 variable on same year variable. So table1 has say key1 cases A and B on year 2000 that are linked to one and same key2 variable in table2. So in effect, I am interested in key1 doubles of a single key2 on specific year.

SELECT query.key2
FROM (
    SELECT DISTINCT a.key1, b.key2, a.year
    FROM table1 AS a JOIN table2 AS b ON a.key1=b.key1
    WHERE a.year IS NOT NULL
    )
    AS query
GROUP BY query.key2, query.year
HAVING COUNT(*)>1

This is the code I have used to get the list of key2 cases that have more than 1 key1 case linked to it for same year. To get the list of all key1 cases linked to these key2 cases it is simple to add that as a subquery for a query on table2. But that results with ALL the key1 cases linked to that specific key2. What I want are specifically the same year doubles. I am dumbstruck at this seemingly easy task, I can't have key in the SELECT because it is not listed in the group by clause and using key2 cases in subquery again won't filter out that non-wanted key1 cases from different years. Any ideas?

Answer 1

Notulysses, that doesn't quite do the trick. It manages to group key1 and key2 cases under desired required year but links too many key1 cases, ie. linking key1 to key2 becomes unconnected of year.

Coder of code: table1 would be like

key 1 year plus bunch of uninteresting variables here

Table 2:

key1 key2 plus few other uninteresting variables.

And what I essentially want is a list of such key1 when key1 cases A and B, both in year 2000, are linked to one key2. Badly explained, I know, but English is not my primary language and without visual aid the idea is not the simplest to convey I find when guys don't know the tables and data.

Anyway, I managed to do the trick. What I essentially did was first create temp table where I saved all key2 cases that have more than one case linked to it for one year. Then II used DENSE_RANK-function and flagged year,key2 doubles with it and rest was easy. As such:

CREATE TABLE #temp2 (key2 VARCHAR(50), key1 VARCHAR(50), year DATE, dr INT)
INSERT INTO #temp2
SELECT b.key2, a.key1, a.year,
DENSE_RANK() OVER (ORDER BY b.key2, a.year DESC) AS dr
FROM table1 AS a JOIN table2 AS b ON a.key1=b.key1
WHERE b.key2 IN (SELECT key2 FROM #temp1) --#temp1 being table with key2's from previous query
ORDER BY b.key2, a.year DESC

SELECT *
FROM #temp2
WHERE dr IN (
    SELECT dr
    FROM #temp2
    GROUP BY dr
    HAVING COUNT(*)>1)

It is a clumsy solution but it works. Better ideas are welcomed!

Answer 2

SELECT t1.*
     , t2.key1
FROM ( 
    SELECT DISTINCT  b.key2
         , a.year
    FROM table1 a 
    LEFT JOIN table2 b ON a.key1 = b.key1
    WHERE b.key1 IS NOT NULL
    GROUP BY b.key2
           , a.year
    HAVING COUNT(a.key1) > 1) t1
JOIN table2 t2 ON t1.key2 = t2.key2

Answer 3

WITH TABLECTE 
AS
    SELECT DISTINCT a.key1, b.key2, a.year
    FROM   table1 AS a JOIN table2 AS b ON a.key1=b.key1
    WHERE  a.year IS NOT NULL
SELECT TABLECTE.key2 
FROM   TABLECTE
GROUP BY TABLECTE .key2, TABLECTE .year
HAVING COUNT(*)>1