PostgreSQL 游标更新表



我有一个代表患者体检的表格,它有检查的ID和患者ID。

我想逐行浏览表格,并获取每个患者 ID 并比较其不同的咨询,看看它是否被视为"new_attack"。我正在处理疟疾疾病,我们认为每个在过去 60 天内接受过咨询并且检测呈阳性的患者都是旧病例和new_attack = false,否则new_attack = true

但是当我浏览表格时,我必须考虑列palufalci,因为患者可以进行体检,但疟疾检测不呈阳性,在这种情况下new_attack = false.

以下是创建表的代码:

CREATE TABLE public.tbl_diagnostiques_guy (
  id integer NOT NULL DEFAULT nextval('tbl_diagnostiques_guy_id_seq'::regclass),
  dateconsultation date,
  numeropatient character varying(13),
  palufalci boolean,
  new_attack boolean
);

我使用此查询来计算每位患者的 2 次体检之间的datediff

SELECT id, numeropatient, palufalci,
    dateconsultation, NextDate,
    date(NextDate) - date(dateconsultation) as Diff, new_attack
FROM (
  SELECT id, numeropatient, palufalci, dateconsultation, new_attack,
         (SELECT  MIN(dateconsultation) 
          FROM    tbl_diagnostiques_guy T2
          WHERE   T2.numeropatient = T1.numeropatient
          AND     T2.dateconsultation > T1.dateconsultation
         ) AS NextDate
  FROM tbl_diagnostiques_guy T1) AS T
WHERE NextDate IS NOT NULL AND (date(NextDate) - date(dateconsultation) < 60)
GROUP BY id, numeropatient, palufalci, dateconsultation, NextDate, new_attack
ORDER BY numeropatient DESC;

结果是:这里

现在我想知道如何更新表并获得我想要的结果。

从您的问题来看,您想用值填充new_attack列。这很容易通过您的查询完成 - 尽管您设置new_attack值的逻辑似乎缺失 - 但实际上有一个使用 lag() 窗口函数的更优雅的形式:

SELECT id, numeropatient, palufalci, dateconsultation,
       CASE WHEN days IS NULL OR days > 60 THEN false
       ELSE palufalci AND old_test
       END AS new_attack
FROM (
  SELECT id, numeropatient, palufalci, lag(palufalci) OVER w AS old_test,
         dateconsultation, dateconsultation - lag(dateconsultation) OVER w AS days,
  FROM tbl_diagnostiques_guy
  WINDOW w AS (PARTITION BY numeropatient ORDER BY dateconsultation) ) sub;

运行该语句以验证它是否按预期执行。如果满意,那么您可以轻松地将整个事情重新加工成UPDATE语句:

UPDATE tbl_diagnostiques_guy t
SET new_attack =
       CASE WHEN days IS NULL OR days > 60 THEN false
       ELSE palufalci AND old_test
       END
FROM (
  SELECT id, numeropatient, palufalci, lag(palufalci) OVER w AS old_test,
         dateconsultation, dateconsultation - lag(dateconsultation) OVER w AS days,
  FROM tbl_diagnostiques_guy
  WINDOW w AS (PARTITION BY numeropatient ORDER BY dateconsultation) ) sub
WHERE t.id = sub.id; -- add other join conditions as required

让我们更详细地看一下new_attack逻辑:

CASE WHEN days IS NULL OR days > 60 THEN false
ELSE palufalci AND old_test
END

第一行WHEN days IS NULL OR days > 60 THEN false表示:如果上一次咨询是 60 多天前的,或者这是第一次咨询(lag()函数会返回第一次咨询的NULL,因为没有前一行),则new_attack值为 false

第二行palufalci AND old_test的意思是:如果在 60 天内进行了第二次咨询,那么只有在之前和当前的测试都true的情况下new_attacktrue

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