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我对 SQL Server 2008 还很陌生,但遇到了问题。我有一个表,其中包含一个带有实验室结果的 varchar(255) 列。列中的数据如下所示:

result_value
7.0
162
NEGATIVE
SEE NOTE
9.6

现在,我要做的是运行以下查询:

;WITH PatientData (patient_id,last_name, first_name, ethnic_group,
race, icdcode, encounter_date, test_performed, result_value,
result_units )
AS
(
SELECT pat.patient_id, pat.last_name, Pat.first_name, pat.ethnic_group,
pat.race, pl.icdcode, enc.encounter_date, lab.test_performed, lab.result_value,
lab.result_units
FROM dbo.dem_patient pat
RIGHT JOIN dbo.med_problemlist as pl
ON pat.patient_id = pl.patient_id
JOIN dbo.enc_encounter as enc
ON pat.patient_id = enc.patient_id
JOIN dbo.med_labresult as lab
ON pat.patient_id = lab.patient_id
WHERE pl.icdcode like '250%' AND
enc.encounter_date >= '01/01/2012' AND enc.encounter_date <= '12/31/2012' AND
pat.last_name != 'test' AND (lab.test_performed = 'HgbA1c' OR 
lab.test_performed = 'Hemoglobin A1c')
GROUP BY pat.patient_id, pat.last_name, Pat.first_name, pat.ethnic_group,
pat.race, pl.icdcode, enc.encounter_date, lab.test_performed, lab.result_value,
lab.result_units
)
select sum(case when result_value < 7.0 then 1 else 0 end) as [LessThan7%],
       sum(case when result_value >= 7.0 and result_value < 8.0 then 1 else 0 end) as [7%to8%],
       sum(case when result_value >= 8.0 and result_value < 9.0 then 1 else 0 end) as [8%to9%],
       sum(case when result_value >= 9.0 then 1 else 0 end) as [GreaterThan9%]
FROM PatientData pd1 
WHERE EXISTS (SELECT patient_id 
                FROM PatientData pd2 
                WHERE pd2.patient_id  = pd1.patient_id  
                GROUP BY patient_id 
                HAVING COUNT(*)>1 )

由于我应该只返回进行了 A1c 测试的患者,因此 results_value 列应该始终看起来像 xx - 但是,我尝试运行此查询时,我得到“将 varchar 转换为数据类型 numeric 的算术溢出错误。

虽然我意识到将这些不同类型的数据放在同一列中并不理想,但我并没有创建数据库——我只需要使用它。反正有这个问题吗?

谢谢您的帮助。

4

1 回答 1

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您可以使用这种解决方案。让我们考虑这个表和数据。

CREATE TABLE [tb](
[id] [int] IDENTITY(1,1) NOT NULL primary key,
[result] [varchar](50) NULL,
) ON [PRIMARY] 
INSERT [tb] ([id], [result]) VALUES (1, N'12')
INSERT [tb] ([id], [result]) VALUES (2, N'24')
INSERT [tb] ([id], [result]) VALUES (3, N'35')
INSERT [tb] ([id], [result]) VALUES (4, N'NEGATIVE')
INSERT [tb] ([id], [result]) VALUES (5, N'SEE NOTE')
INSERT [tb] ([id], [result]) VALUES (6, N'15.5')
INSERT [tb] ([id], [result]) VALUES (7, N'16.4')

您可以使用 ASCII 函数(ASCII 返回字符表达式的最左侧字符的 ASCII 码值。)从此示例数据中获取正确的数字

 select id,ascii(result),convert(decimal(16,2),result) as result from tb
 where ascii(result)>=ascii('0') and ascii(result)<=ascii('9')
  union 
 select id,ascii(result),-1.0 as result from tb
 where ascii(result)<=ascii('0') or ascii(result)>=ascii('9')

您可以根据需要标记无效记录。您可以向此查询添加更多逻辑。您可以将其用作子查询或与“join”一起使用。我希望它会帮助你。

马辛·帕兹吉尔

于 2013-03-05T20:45:59.400 回答