同一个表的两个行值的差异

Question

我有一个实验室值表，我想查询并获得最后一个值和最近值之间的差异，如果它的＆gt; = 0.2。我该怎么做呢。这是我到目前为止的查询。任何帮助，将不胜感激。数据库是Oracle 10g。有关详细信息，请参见屏幕截图。

SELECT DISTINCT VISITLAB.DATEOFENTRY, VISITLAB.LABVALUE, VISITLAB.CODE      
FROM(XCX.PATIENTVISIT PATIENTVISIT
       INNER JOIN
          XCX.MASTERPATIENT MASTERPATIENT
       ON (PATIENTVISIT.MASTERPATIENTID = MASTERPATIENT.MASTERPATIENTID))
   INNER JOIN
      XCX.VISITLAB VISITLAB
   ON (VISITLAB.MASTERPATIENTID = PATIENTVISIT.MASTERPATIENTID)
      AND (VISITLAB.VISITNUMBER = PATIENTVISIT.VISITNUMBER)  
WHERE     (MASTERPATIENT.MASTERPATIENTID = 'xxxxxxxx')
   AND (VISITLAB.CODE = 'UQN0')
   AND (PATIENTVISIT.DISCHARGEDATE IS NULL)
   AND (PATIENTVISIT.FACILITYCODE = 'x')

Answer 1

您应该使用LAG anayltical函数来检索上一行中的值。

SELECT * FROM (
SELECT DISTINCT VISITLAB.DATEOFENTRY, VISITLAB.LABVALUE, VISITLAB.CODE, 
    (LAG(VISITLAB.LABVALUE) over (order by VISITLAB.DATEOFENTRY) - VISITLAB.LABVALUE) as DIFF
FROM(XCX.PATIENTVISIT PATIENTVISIT
INNER JOIN
       XCX.MASTERPATIENT MASTERPATIENT
       ON (PATIENTVISIT.MASTERPATIENTID = MASTERPATIENT.MASTERPATIENTID))
INNER JOIN
       XCX.VISITLAB VISITLAB
       ON (VISITLAB.MASTERPATIENTID = PATIENTVISIT.MASTERPATIENTID)
       AND (VISITLAB.VISITNUMBER = PATIENTVISIT.VISITNUMBER)
WHERE (MASTERPATIENT.MASTERPATIENTID = 'xxxxxxxx')
AND (VISITLAB.CODE = 'UQN0')
AND (PATIENTVISIT.DISCHARGEDATE IS NULL)
AND (PATIENTVISIT.FACILITYCODE = 'x')
) INLINEVIEW
WHERE DIFF < .2

这不一定是您正在寻找的答案，但它演示了如何根据前一行中的值计算派生值。

Answer 2

对于给定的CODE（例如UQN0），是否需要跨所有LABVALUE计算这些DIFF，这一点并不十分清楚。如果您对发布的屏幕截图中显示的最后两个感兴趣，那么也可以使用没有任何分析函数的纯SQL：

create table lab (
  code varchar2(10), 
  date_of_entry date, 
  lab_value number(10,2)
);
insert into lab values('UQN0', trunc(sysdate) - 3, 0.98);
insert into lab values('UQN0', trunc(sysdate) - 2, 0.85);
insert into lab values('UQN0', trunc(sysdate) - 1, 0.90);
insert into lab values('UQN0', trunc(sysdate), 0.64);
commit;
select * from lab;

with last_results as (
  select t.* from (
    select * from lab where code = 'UQN0' 
   order by date_of_entry desc
  ) t where rownum <= 2
)
select t1.*
from last_results t1, last_results t2 
  where t1.lab_value - t2.lab_value  >= 0.2;

如果不关心确保与其他数据库服务器的兼容性，那么LAG（）函数显然是可行的方法。