在Oracle SQL中向後Oracle INSTR

Question

我想在oracle SQL中找到一種方法，將第一個「/」從右到第四個「/」之後的所有文本。我需要做的這一個SQL語句..

table name: access_log 
col name: download 
value: Download file:/webdocs/data/groupXXX/case/03_28_54_9_0000011856.pdf 

我試圖用

case 

那麼有多少匹配的計數結束了？這可以做到嗎？

Answer 1

一個簡單SUBSTR方法從Oracle 10g的+

select TRIM(BOTH '/' FROM 
      substr(
        download, 
        INSTR(download,'/',1,4), 
        INSTR(download,'/',1,5)-INSTR(download,'/',1,4) 
       ) 
      ) 
FROM ACCESS_LOG; 

作品和計數可能是，

SELECT 
COUNT(
     CASE WHEN TRIM(BOTH '/' FROM 
         substr(
           download, 
           INSTR(download,'/',1,4), 
           INSTR(download,'/',1,5)-INSTR(download,'/',1,4) 
           ) 
         ) = 'case' 
     THEN 
      1 
     END 
     ) COUNT_MATCHED 
FROM 
    ACCESS_LOG; 

Answer 2

select regexp_replace('Download file:/webdocs/data/groupXXX/case/03_28_54_9_0000011856.pdf', 
         '.*/([^/]+)/[^/]*$', '\1') from dual;