1

I only found slightly different examples I couldn't adapt to my needs due to my very limited sql skills.

I have a table with 3 revelant columns:

ItemID  Date      Result
1       1.2.2014  A
5       6.4.2014  B
9       7.4.2014  A
1       8.4.2014  A
1       9.4.2014  A
1       10.4.2014 A

I want to find the Items that had a particular result (let's say A) 3 times consecutively. In the sample above it would be Item 1. The dates are not normally consecutive.

It should work in Oracle SQL.

Many thanks for the help!

4

3 回答 3

1 
SQL> WITH DATA AS(
  2  SELECT 1 ITEM_ID, TO_DATE('1.2.2014','DD.MM.YYYY') DT, 'A' RSLT FROM DUAL UNION ALL
  3  SELECT 5,       TO_DATE('6.4.2014','DD.MM.YYYY')  , 'B' RSLT FROM DUAL UNION ALL
  4  SELECT 9,       TO_DATE('7.4.2014','DD.MM.YYYY')  , 'A' RSLT FROM DUAL UNION ALL
  5  SELECT 1,       TO_DATE('8.4.2014','DD.MM.YYYY')  , 'A' RSLT FROM DUAL UNION ALL
  6  SELECT 1,       TO_DATE('9.4.2014','DD.MM.YYYY')  , 'A' RSLT FROM DUAL UNION ALL
  7  SELECT 1,       TO_DATE('10.4.2014','DD.MM.YYYY') , 'A' RSLT FROM DUAL)
  8  SELECT ITEM_ID FROM(
  9  SELECT A.*, ROW_NUMBER() OVER(PARTITION BY ITEM_ID ORDER BY RSLT) RN
 10  FROM DATA A)
 11  WHERE RN =3
 12  /

   ITEM_ID
----------
         1

SQL>
于 2014-09-30T09:22:02.150 回答
0

Not sure if this is the best way of of achieving it, but it works. I created the following table to put your example data into:

CREATE TABLE DATE_RESULT(
  ITEM_ID INT,
  DATE_COL DATE,
  RESULT_COL VARCHAR2(255 CHAR)
);

Then ran this query:

SELECT ITEM_ID FROM(
  SELECT
    ITEM_ID,
    TO_CHAR(DATE_COL,'DD-MON-YYYY') AS DATE_COL,
    RESULT_COL,
    LAG(ITEM_ID,1,NULL) OVER (ORDER BY ITEM_ID ASC, DATE_COL ASC) AS LAST_ITEM_ID,
    LAG(ITEM_ID,2,NULL) OVER (ORDER BY ITEM_ID ASC, DATE_COL ASC) AS LAST_ITEM_ID2,
    LAG(TO_CHAR(DATE_COL,'DD-MON-YYYY'),1,NULL) OVER (ORDER BY ITEM_ID ASC, DATE_COL ASC) AS LAST_DATE,
    LAG(TO_CHAR(DATE_COL,'DD-MON-YYYY'),2,NULL) OVER (ORDER BY ITEM_ID ASC, DATE_COL ASC) AS LAST_DATE2,
    LAG(RESULT_COL,1,NULL) OVER (ORDER BY ITEM_ID ASC, DATE_COL ASC) AS LAST_RESULT,
    LAG(RESULT_COL,2,NULL) OVER (ORDER BY ITEM_ID ASC, DATE_COL ASC) AS LAST_RESULT2
  FROM 
    DATE_RESULT
)
WHERE 
  ITEM_ID = LAST_ITEM_ID
  AND ITEM_ID = LAST_ITEM_ID2
  AND TO_DATE(DATE_COL)-1 = TO_DATE(LAST_DATE)
  AND TO_DATE(DATE_COL)-2 = TO_DATE(LAST_DATE2)
  AND RESULT_COL = LAST_RESULT
  AND RESULT_COL = LAST_RESULT2;

The query uses Oracle's LAG() function to get the values from previous rows. So in this example, LAST_ITEM_ID is the item ID from the previous row, and LAST_ITEM_ID is the item ID from 2 rows previous.

In the WHERE clause I make sure that the ITEM_ID matches the previous two ITEM_IDs and that the RESULT_COL matches the previous two RESULT_COLs. I also make sure that the last two dates were consecutive.

于 2014-09-30T09:21:36.043 回答
0

Thanks to user1578653 I figured it out. Maybe there is a more elegant way to do it, but it worked for about 500000 records in a few seconds:

SELECT ITEM_ID FROM(
   SELECT ITEM_ID, DATECOL, RESULT,
     LAG(ITEM_ID,1,NULL) OVER (ORDER BY ITEM_ID ASC, DATECOL ASC) AS LASTITEM,
     LAG(ITEM_ID,2,NULL) OVER (ORDER BY ITEM_ID ASC, DATECOL ASC) AS LASTITEM2,
     LAG(RESULT,1,NULL) OVER (ORDER BY ITEM_ID ASC, DATECOL ASC) AS LASTCODE,
     LAG(RESULT,2,NULL) OVER (ORDER BY ITEM_ID ASC, DATECOL ASC) AS LASTCODE2
   FROM
     RESULTTABLE
   )
WHERE
   ITEM_ID = LASTITEM
   AND ITEM_ID = LASTITEM2
   AND RESULT = 'A'
   AND RESULT = LASTCODE
   AND RESULT = LASTCODE2;

I was looking for a result of 'A' that occured 3 times consecutively in this example.

I'm rather happy, thanks!

于 2014-09-30T13:36:09.170 回答