如何从字符串中提取子字符串？

Question

fnd YGY LOOKUP_TYPE = 'welcome' HELO HIASDH LOOKUP_TYPE = 'home' hello how are you? 

以上是字符串，我想输出

welcome 
home 

Answer 1

查询1：

注：单引号需要在文字文本进行转义：

WITH test_data (string) AS (
    SELECT 'LOOKUP_TYPE = ''welcome'' LOOKUP_TYPE = ''home''' 
    FROM DUAL 
) 
SELECT SUBSTR(string, quote1 + 1, quote2 - quote1 - 1) AS first_quoted_string, 
     SUBSTR(string, quote3 + 1, quote4 - quote3 - 1) AS second_quoted_string, 
FROM (
    SELECT string, 
     INSTR(string, '''', 1, 1) AS quote1, 
     INSTR(string, '''', 1, 2) AS quote2, 
     INSTR(string, '''', 1, 3) AS quote3, 
     INSTR(string, '''', 1, 4) AS quote4 
    FROM test_data 
) 

输出：

FIRST_QUOTED_STRING SECOND_QUOTED_STRING 
------------------- -------------------- 
welcome    home 

查询2：

注：另类text literal报价demiliters q'[your string]'使用，使得单引号不需要进行转义：

WITH test_data (string) AS (
    SELECT q'[LOOKUP_TYPE = 'welcome' LOOKUP_TYPE = 'home']' 
    FROM DUAL 
), 
quotes (string, quote_start, quote_end, lvl) AS (
    SELECT string, 
     INSTR(string, '''', 1, 1), 
     INSTR(string, '''', 1, 2), 
     1 
    FROM test_data 
    WHERE INSTR(string, '''', 1, 2) > 0 
UNION ALL 
    SELECT string, 
     INSTR(string, '''', 1, 2 * lvl + 1), 
     INSTR(string, '''', 1, 2 * lvl + 2), 
     lvl + 1 
    FROM quotes 
    WHERE INSTR(string, '''', 1, 2 * lvl + 2) > 0 
) 
SELECT SUBSTR(string, quote_start + 1, quote_end - quote_start - 1) AS quoted_string 
FROM quotes 

输出：

QUOTED_STRING 
------------- 
welcome 
home 

查询3 - 正则表达式：

注：用于获取输入:your_string绑定变量（你也可以使用文本文字代替）：

WITH test_data (string) AS (
    SELECT :your_string FROM DUAL 
), 
quotes (string, quoted_string, lvl, max_lvl) AS (
    SELECT string, 
     REGEXP_SUBSTR(string, q'[LOOKUP_TYPE\s*=\s*("|''?)(.*?)\1]', 1, 1, NULL, 2), 
     1, 
     REGEXP_COUNT(string, q'[LOOKUP_TYPE\s*=\s*("|''?)(.*?)\1]') 
    FROM test_data 
    WHERE REGEXP_COUNT(string, q'[LOOKUP_TYPE\s*=\s*("|''?)(.*?)\1]') > 0 
UNION ALL 
    SELECT string, 
     REGEXP_SUBSTR(string, q'[LOOKUP_TYPE\s*=\s*("|''?)(.*?)\1]', 1, lvl + 1, NULL, 2), 
     lvl + 1, 
     max_lvl 
    FROM quotes 
    WHERE lvl < max_lvl 
) 
SELECT quoted_string 
FROM quotes 

输出：

QUOTED_STRING 
------------- 
welcome 
home 

Answer 2

with t (col) as (select q'[fnd YGY LOOKUP_TYPE = 'welcome' HELO HIASDH LOOKUP_TYPE = "home" hello LOOKUP_TYPE = ''man'' how are you?]' from dual) 

select rtrim(regexp_replace(col,q'[(.*?LOOKUP_TYPE\s*=\s*(''|'|")(.*?)\2|.+)]','\3,'),',') as lookup_types 
from t 

---

+------------------+ 
| LOOKUP_TYPES | 
+------------------+ 
| welcome,home,man | 
+------------------+