-12

次のようなテーブルがあります。

id name center1    center2   cnter3
1  abc  baroda     bhopal    chennai
2  pqr  amhedabad  Surat     kolkata
3  nml  bhopal     chennai   Surat
4  fts  baroda     Surat     Baroda

次のような結果が欲しい：

         center1 center2  center3
Baroda     2       0        1   
Ahmedabad  1       0        0
bhopal     1       1        0
chennai    0       1        1
Kolkatta   0       0        1
Surat      0       2        1.
4

2 に答える 2

3

This will require a handful of subquery joins. First, you need to get all the distinct values from the three columns, which would be done with a UNION of queries of each column.

That distinct list must then be LEFT JOINed against a trio of subqueries which pulls the counts per value from each of the three columns.

Finally, a COALESCE() replaces NULLs with 0.

SELECT 
  allvalues.value,
  /* LEFT JOIN will result in NULL for non-existent values, so coalesce into 0 */
  COALESCE(c1.counts, 0) AS center1,
  COALESCE(c2.counts, 0) AS center2,
  COALESCE(c3.counts, 0) AS center3
FROM
  (
    /* UNIONs gets distinct vals from all columns */
    SELECT center1 AS value FROM yourtable
    UNION
    SELECT center2 AS value FROM yourtable
    UNION
    SELECT center3 AS value FROM yourtable
  ) AS allvalues
  /* LEFT JOIN the list of distinct possible values against a count per value for each column */
  LEFT JOIN (
    SELECT center1, COUNT(*) AS counts FROM yourtable GROUP BY center1
  ) c1 ON allvalues.value = c1.center1
  LEFT JOIN (
    SELECT center2, COUNT(*) AS counts FROM yourtable GROUP BY center2
  ) c2 ON allvalues.value = c2.center2
  LEFT JOIN (
    SELECT center3, COUNT(*) AS counts FROM yourtable GROUP BY center3
  ) c3 ON allvalues.value = c3.center3

Here is a working demonstration, because I guess I'm feeling extra charitable tonight.

于 2012-09-11T01:32:13.383 に答える
2

sqlfiddleを使用して、テーブルのテキスト表現を一連の CREATE TABLE および INSERT INTO ステートメントに変換しました。

CREATE TABLE CenterCodes
    (`id` int, `name` varchar(3), `center1` varchar(9), `center2` varchar(7), `center3` varchar(7))
;

INSERT INTO CenterCodes
    (`id`, `name`, `center1`, `center2`, `center3`)
VALUES
    (1, 'abc', 'baroda', 'bhopal', 'chennai'),
    (2, 'pqr', 'amhedabad', 'Surat', 'kolkata'),
    (3, 'nml', 'bhopal', 'chennai', 'Surat'),
    (4, 'fts', 'baroda', 'Surat', 'Baroda')
;

次のクエリを試してください。内側UNIONは、文字列値ごとに複数の行を生成し、各セットの 3 つの列のうち 2 つが NULL です。これらの NULL はSUM()、外側のクエリの集計によって削除され、行セットが文字列値ごとに 1 行に縮小されます。

SELECT
  centername,
  SUM(CASE WHEN centername = center1 THEN 1 ELSE 0 END) AS center1,
  SUM(CASE WHEN centername = center2 THEN 1 ELSE 0 END) AS center2,
  SUM(CASE WHEN centername = center3 THEN 1 ELSE 0 END) AS center3
FROM (
  SELECT
    center1 AS centername,
    center1,
    NULL AS center2,
    NULL AS center3
  FROM CenterCodes
  UNION ALL
  SELECT
    center2 AS centername,
    NULL AS center1,
    center2,
    NULL AS center3
  FROM CenterCodes
  UNION ALL
  SELECT
    center3 AS centername,
    NULL AS center1,
    NULL AS center2,
    center3
  FROM CenterCodes
) AS Centers
GROUP BY centername;
于 2012-09-11T01:42:08.563 に答える