SQL - 计算多列中的出现次数

Question

我有一个假设的数据库，其中包含患者及其合并症（或通常称为诊断代码）：

PATIENT_ID | HOSPITAL | DIAG_01 | DIAG_02 | DIAG_03 | DIAG_04 ...up to Diag_14
 Patient 1 |  Hosp1   |  C01    |  C02    |  NULL   |  NULL
 Patient 2 |  Hosp1   |  A01    |  A02    |  A03    |  P37
 Patient 3 |  Hosp2   |  A01    |  D10    |  X01    |  NULL

患者的合并症（最多可记录 14 种）定义了患者可能患有的任何疾病/疾病。特别是 C00 - C97 和 D00 - D48 与癌症 ICD-10 代码相关。

我想为每家医院制作一份摘要，计算有多少患者的诊断列中有癌症 ICD-10 代码。

根据上述示例，Hosp1 和 Hosp2 各有 1 名癌症患者。患者 1 的 C01 和 C02 诊断代码将仅计为一名癌症患者。但是，如果患者 1 再次入院，则将被视为另一例癌症患者。

我尝试创建一个包含所有癌症合并症的表，并对所有 14 个诊断列使用 JOIN 查询，但无法进行计数。

任何想法或帮助将不胜感激。

Answer 1

全部使用联合

SELECT * FROM /**customize your select here could be count() etc **/
 (SELECT PATIENT_ID,HOSPITAL,DIAGNOSIS
FROM (
    SELECT PATIENT_ID, HOSPITAL, LEFT(DIAG_01,1) AS DIAGNOSIS FROM table_name WHERE DIAG_01 IS NOT NULL UNION ALL   
    SELECT PATIENT_ID, HOSPITAL, LEFT(DIAG_02,1) AS DIAGNOSIS FROM table_name WHERE DIAG_02 IS NOT NULL UNION ALL
    SELECT PATIENT_ID, HOSPITAL, LEFT(DIAG_03,1) AS DIAGNOSIS FROM table_name WHERE DIAG_03 IS NOT NULL UNION ALL
    SELECT PATIENT_ID, HOSPITAL, LEFT(DIAG_04,1) AS DIAGNOSIS FROM table_name WHERE DIAG_04 IS NOT NULL
      ) AS T1
GROUP BY PATIENT_ID,HOSPITAL,DIAGNOSIS
) AS T2
WHERE DIAGNOSIS IN ('C','D') /**where **/

Answer 2

您的数据布局错误。你应该有一个连接表。但是，你没有。你可以用union all做你想做的事：

select hospital, count(distinct patient_id)
from ((select patient_id, hospital, diag_01 as diag
       from hypothetical
      ) union all
      (select patient_id, hospital, diag_02 as diag
       from hypothetical
      )
      union all
      . . .
      (select patient_id, hospital, diag_14 as diag
       from hypothetical
      )
     ) phd
where diag in ('C01', 'C02');