去字段中的内容

来源：互联网发布：四层别墅网络覆盖方案编辑：程序博客网时间：2024/05/01 08:37

处理年龄问题

AGE AGE

007Y 7岁

045D 45天

013M 13月

SELECT
       case
          when ascii(SUBSTR(replae(patientvisit.age,' ',''), 1, 1)) >= '48' and
               ascii(SUBSTR(patientvisit.age, 1, 1)) <= '57' then
           SUBSTR(patientvisit.age, 1, 3) || (case
          when SUBSTR(patientvisit.age, 4, 1) LIKE 'Y' then
           '岁'
          when SUBSTR(patientvisit.age, 4, 1) LIKE 'M' then
           '月'
          when SUBSTR(patientvisit.age, 4, 1) LIKE 'D' then
           '天'
        END)
        WHEN ascii(SUBSTR(patientvisit.age, 2, 1)) >= '48'and ascii(SUBSTR(patientvisit.age, 2, 1)) <= '57'
        THEN SUBSTR(patientvisit.age, 2, 2) || (case
         when SUBSTR(patientvisit.age,
                     4,
                     1) LIKE 'Y' then
          '岁'
         when SUBSTR(patientvisit.age,
                     4,
                     1) LIKE 'M' then
          '月'
         when SUBSTR(patientvisit.age,
                     4,
                     1) LIKE 'D' then
          '天'
       END)
       ELSE SUBSTR(patientvisit.age, 3, 1) || (case
         when SUBSTR(patientvisit.age,
                     4,
                     1) LIKE 'Y' then
          '岁'
         when SUBSTR(patientvisit.age,
                     4,
                     1) LIKE 'M' then
          '月'
         when SUBSTR(patientvisit.age,
                     4,
                     1) LIKE 'D' then
          '天'
       END) END AS Age
from patientvisit