Question

我试图从字符串中解析一个美元金额。

示例字符串：

* SOC 1369.00 - NCS 1239.46 = PT LIAB 129.54
* SOC 1369.00 - NCS 1239.46 = PT LIAB 140
* SOC = 1178.00
* SOC 1622.00 - NCS 209.74 = PT LIAB 1412.26 RECIPIENT AGE
* LINE＃1 SOC 0.00 - NCS 22.77 = LIAB -22.77
SOC MET AND CLEARED，SOC 2062-NCS 498.56 = PT LIABLE 1563.44
* SOC 1622.00 - NCS 209.74 = PT LIAB 1412 RECIPIENT AGE 1234

我想提取患者责任，即文字后面的美元金额＆＃34; LIAB，＆＃34; ＆＃34; PT LIAB，＆＃34;或者＆＃34; LIABLE。＆＃34;美元金额可以是负数，可能有也可能没有小数。

我的解决方案：

    REPLACE(REPLACE(REGEXP_SUBSTR(REMARKS,'LIAB+[LE]?+ (-?+\d+[.]?)+\d'),'LIAB ',''),'LIABLE ','')

这似乎有点笨拙，我认为这是一个更简单的解决方案。任何指导意见将不胜感激！

我使用Toad for Oracle 12.8。

Answer 1

尝试一下：

SQL> with tbl(rownbr, remarks) as (
         select 1, '*SOC 1369.00 - NCS 1239.46 = PT LIAB 129.54'                from dual union
         select 2, '*SOC 1369.00 - NCS 1239.46 = PT LIAB 140'                   from dual union
         select 3, '*SOC = 1178.00'                                             from dual union
         select 4, '*SOC 1622.00 - NCS 209.74 = PT LIAB 1412.26 RECIPIENT AGE'  from dual union
         select 5, '*LINE #1 SOC 0.00 - NCS 22.77 = LIAB -22.77'                from dual union
         select 6, 'SOC MET AND CLEARED, SOC 2062-NCS 498.56=PT LIABLE 1563.44' from dual union
         select 7, '*SOC 1622.00 - NCS 209.74 = PT LIAB 1412 RECIPIENT AGE 1234' from dual
       )
       select rownbr,
             case
               when remarks = regexp_replace(remarks, '.*((LIAB|LIABLE) ([-.0-9]+)).*$', '\3') then
                 '0' -- regexp_replace returns the orig string if the pattern is not found.
               else
                 regexp_replace(remarks, '.*((LIAB|LIABLE) ([-.0-9]+)).*$', '\3')
             end patient_liability
      from tbl;

    ROWNBR PATIENT_LIABILITY
---------- -------------------------
         1 129.54
         2 140
         3 0
         4 1412.26
         5 -22.77
         6 1563.44
         7 1412

7 rows selected.

SQL>

Answer 2

您几乎可以使用REGEXP_REPLACE()和后退参考：

REGEXP_REPLACE(REMARKS,'.*(PT LIAB|LIAB|LIABLE) (-?\d+[.]?\d+).*', '\2')

...但是通过一个没有匹配模式的值来传递（因此你的第三个例子仍然会得到*SOC = 1178.00。你可以使用一个案例表达式和REGEXP_LIKE()来避免这种情况：

with t (remarks) as (
  select '*SOC 1369.00 - NCS 1239.46 = PT LIAB 129.54' from dual
  union all select '*SOC 1369.00 - NCS 1239.46 = PT LIAB 140' from dual
  union all select '*SOC = 1178.00' from dual
  union all select '*SOC 1622.00 - NCS 209.74 = PT LIAB 1412.26 RECIPIENT AGE' from dual
  union all select '*LINE #1 SOC 0.00 - NCS 22.77 = LIAB -22.77' from dual
  union all select 'SOC MET AND CLEARED, SOC 2062-NCS 498.56=PT LIABLE 1563.44' from dual
)
SELECT REMARKS,
  CASE WHEN REGEXP_LIKE(REMARKS, '.*(PT LIAB|LIAB|LIABLE) (-?\d+[.]?\d+).*')
    THEN REGEXP_REPLACE(REMARKS,'.*(PT LIAB|LIAB|LIABLE) (-?\d+([.]\d+)?).*', '\2')
    END as liability
from t;

REMARKS                                                    LIABILITY
---------------------------------------------------------- ----------
*SOC 1369.00 - NCS 1239.46 = PT LIAB 129.54                129.54    
*SOC 1369.00 - NCS 1239.46 = PT LIAB 140                   140       
*SOC = 1178.00                                                       
*SOC 1622.00 - NCS 209.74 = PT LIAB 1412.26 RECIPIENT AGE  1412.26   
*LINE #1 SOC 0.00 - NCS 22.77 = LIAB -22.77                -22.77    
SOC MET AND CLEARED, SOC 2062-NCS 498.56=PT LIABLE 1563.44 1563.44

但这似乎没有那么好，而且两次使用正则表达式会让它变得更加昂贵。（它可能还可以简化......）。您也可以使用REGEXP_REPLACE()代替两个普通REPLACE()来电：

REGEXP_REPLACE(
  REGEXP_SUBSTR(REMARKS, '(PT LIAB|LIAB|LIABLE) (-?\d+([.]\d+)?)'),
  '(PT LIAB|LIAB|LIABLE) ')

但是这又使它变得更加昂贵。

Oracle REGEXP_SUBSTR解析美元金额

2 个答案: