Question

IMAGE_PATH（其列名）

   sph/images///////30_Fairhall_Court.jpeg
   sph/images///8_Furnival_Court.jpeg 
   sph/images/9_Pennethorne_House.jpeg
   rbkc/images/TAVISTOCK_CRESCENT.jpeg
   haringey///images/399932thumb.jpg
   urbanchoice/images//18190862.jpg
   westminster/images//7_Glarus_Court.jpeg

我想从IMAGE_PATH列中获取子串作为

30_Fairhall_Court
8_Furnival_Court
9_Pennethorne_House
TAVISTOCK_CRESCENT
399932thumb
18190862
7_Glarus_Court

意味着我需要在最后一个＆＃39; /＆＃39;之间使用子字符串。之前＆＃39;。＆＃39; 请给我解决方案如何修复

Answer 1

尝试这种简单的方法（将SUBSTR与INSTR一起使用）：

SELECT SUBSTR( image_path, INSTR(image_path, '/', -1)+1) "Image name"
FROM IMAGES;

编辑：如果您不想使用extension使用.个文件：

SELECT SUBSTR( image_path, INSTR(image_path, '/', -1)+1,
               (INSTR(image_path, '.', -1))-INSTR(image_path, '/', -1)-1) "Image name"
FROM IMAGES;

<强>输出：

30_Fairhall_Court
8_Furnival_Court
9_Pennethorne_House
TAVISTOCK_CRESCENT
399932thumb
18190862
7_Glarus_Court

Answer 2

您可以尝试以下内容：

WITH d1 AS (
    SELECT 'sph/images///////30_Fairhall_Court.jpeg' AS image_path FROM dual
     UNION ALL
    SELECT 'sph/images///8_Furnival_Court.jpeg' AS image_path FROM dual
     UNION ALL
    SELECT 'sph/images/9_Pennethorne_House.jpeg' AS image_path FROM dual
     UNION ALL
    SELECT 'rbkc/images/TAVISTOCK_CRESCENT.jpeg' AS image_path FROM dual
     UNION ALL
    SELECT 'haringey///images/399932thumb.jpg' AS image_path FROM dual
     UNION ALL
    SELECT 'urbanchoice/images//18190862.jpg' AS image_path FROM dual
     UNION ALL
    SELECT 'westminster/images//7_Glarus_Court.jpeg' AS image_path FROM dual
)
SELECT TRIM(TRAILING '.' FROM REGEXP_SUBSTR(image_path, '[^/]+\.'))
  FROM d1;

以上假设最终.之前没有/个字符。更好的解决方案可能是：

WITH d1 AS (
    SELECT 'sph/images///////30_Fairhall_Court.jpeg' AS image_path FROM dual
     UNION ALL
    SELECT 'sph/images///8_Furnival_Court.jpeg' AS image_path FROM dual
     UNION ALL
    SELECT 'sph/images/9_Pennethorne_House.jpeg' AS image_path FROM dual
     UNION ALL
    SELECT 'rbkc/images/TAVISTOCK_CRESCENT.jpeg' AS image_path FROM dual
     UNION ALL
    SELECT 'haringey///images/399932thumb.jpg' AS image_path FROM dual
     UNION ALL
    SELECT 'urbanchoice/images//18190862.jpg' AS image_path FROM dual
     UNION ALL
    SELECT 'westminster/images//7_Glarus_Court.jpeg' AS image_path FROM dual
)
SELECT REGEXP_SUBSTR(image_path, '[^.]+', INSTR(image_path, '/', -1) + 1) 
  FROM d1;

也就是说，在最终.之后找到所有非.（请注意，在字符类中转义/}字符不是必需的。）

输出：

30_Fairhall_Court
8_Furnival_Court
9_Pennethorne_House
TAVISTOCK_CRESCENT
399932thumb
18190862
7_Glarus_Court

我有一个带有IMAGE_PATH列的tableA，我需要从string中存活

2 个答案: