Oracle REGEXP_SUBSTR 解析美元金额

Question

我正在尝试从字符串中解析美元金额。

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 收件人年龄
*第 1 行 SOC 0.00 - NCS 22.77 = LIAB -22.77
SOC 满足并清除，SOC 2062-NCS 498.56=PT 责任 1563.44
*SOC 1622.00 - NCS 209.74 = PT LIAB 1412 收件人年龄 1234

我想拉patient liability，就是dollar amount following text "LIAB," "PT LIAB," or "LIABLE." dollar amount可以是负数，可以有也可以没有一个小数。

我的解决方案：

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

这似乎有点笨拙，我认为有更简单的解决方案。任何指导将不胜感激！

我正在使用 Toad for Oracle 12.8。

Answer 1

你几乎可以通过 REGEXP_REPLACE() 和反向引用到达那里：

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

... 但是它通过未触及的方式传递了一个没有匹配模式的值（所以你的第三个例子仍然会得到 *SOC = 1178.00。你可以使用 case 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+)?).*', '')
    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

但这似乎并没有好多少，并且使用两次正则表达式会使它变得更加昂贵。（它可能还可以简化......）。您还可以使用 REGEXP_REPLACE() 而不是两个普通的 REPLACE() 调用：

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

但这又让它变得更贵了。

Answer 2

试一试：

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]+)).*$', '') then
                 '0' -- regexp_replace returns the orig string if the pattern is not found.
               else
                 regexp_replace(remarks, '.*((LIAB|LIABLE) ([-.0-9]+)).*$', '')
             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>

Oracle REGEXP_SUBSTR 解析美元金额

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