从Oracle SQL中的文本中提取数值

huangapple go评论60阅读模式
英文:

Extract value from a text in oracle SQL

问题

我要翻译的内容:

我有一列,其中包含以下内容

(DESCRIPTION=(ADDRESS=(PROTOCOL=tcps)(HOST="test1.zurich.com")(PORT=1521))
(CONNECT_DATA=(SERVICE_NAME=apps_test1.zurich.com))
(SECURITY=(MY_WALLET_DIRECTORY="/u01/zurich/Fdbb48380hdsvh8/data/zurich_WALLET_DIR")
(SSL_SERVER_DN_MATCH=FALSE)))

现在我只想从完整内容中提取 zurich_WALLET_DIR,如果 MY_WALLET_DIRECTORY 不存在,应该返回 null 值。有人可以帮助我实现吗?

尝试以下但没有成功

SELECT REGEXP_REPLACE(HOST,'(.)(MY_WALLET_DIRECTORY)[[:blank:]]="[[:blank:]]([^)]+)").','\3',1,1,'i')
FROM test;

英文:

I have one column which has the below content

(DESCRIPTION=(ADDRESS=(PROTOCOL=tcps)(HOST="test1.zurich.com")(PORT=1521))
(CONNECT_DATA=(SERVICE_NAME=apps_test1.zurich.com))
(SECURITY=(MY_WALLET_DIRECTORY="/u01/zurich/Fdbb48380hdsvh8/data/zurich_WALLET_DIR")
(SSL_SERVER_DN_MATCH=FALSE)))

Now I want to extract only zurich_WALLET_DIR from complete content and if MY_WALLET_DIRECTORY doesn't exist it should give me null value. Can someone help me achieve this?

Tried below but no success

SELECT REGEXP_REPLACE(HOST,'(.*)(MY_WALLET_DIRECTORY)[[:blank:]]*="[[:blank:]]*([^\)]+*)\"\).*','',1,1,'i') 
  FROM test;

答案1

得分: 1

以下是您要翻译的代码部分:

SELECT REGEXP_SUBSTR(
         HOST,
         '\(MY_WALLET_DIRECTORY\s*=\s*"(.*?)"\)',
         1,
         1,
         'i',
         1
       ) AS wallet_dir
from   test;

或者,更加易于理解的字符串函数方法,速度比正则表达式要快得多:

SELECT CASE
       WHEN start_pos > 0 AND end_pos > 0
       THEN SUBSTR(host, start_pos + 22, end_pos - start_pos - 22)
       END AS wallet_dir
FROM   (
  SELECT host,
         INSTR(host, '(MY_WALLET_DIRECTORY="', 1) AS start_pos,
         INSTR(host, '")', INSTR(host, '(MY_WALLET_DIRECTORY="', 1) + 22)AS end_pos
  from   test
);

对于示例数据:

CREATE TABLE test (host) AS
SELECT '(DESCRIPTION=(ADDRESS=(PROTOCOL=tcps)(HOST="test1.zurich.com")(PORT=1521))(CONNECT_DATA=(SERVICE_NAME=apps_test1.zurich.com))(SECURITY=(MY_WALLET_DIRECTORY="/u01/zurich/Fdbb48380hdsvh8/data/zurich_WALLET_DIR")(SSL_SERVER_DN_MATCH=FALSE)))' FROM DUAL;

两者的输出结果如下:

WALLET_DIR
/u01/zurich/Fdbb48380hdsvh8/data/zurich_WALLET_DIR

如果您只想获取字符串的最后部分:

SELECT REGEXP_SUBSTR(
         HOST,
         '\(MY_WALLET_DIRECTORY\s*=\s*"[^"]*/(.*?)"\)',
         1,
         1,
         'i',
         1
       ) AS wallet_dir
from   test;

或者:

SELECT SUBSTR(wallet_dir, INSTR(wallet_dir, '/', -1) + 1) AS wallet_dir
FROM   (
  SELECT CASE
         WHEN start_pos > 0 AND end_pos > 0
         THEN SUBSTR(host, start_pos + 22, end_pos - start_pos - 22)
         END AS wallet_dir
  FROM   (
    SELECT host,
           INSTR(host, '(MY_WALLET_DIRECTORY="', 1) AS start_pos,
           INSTR(host, '")', INSTR(host, '(MY_WALLET_DIRECTORY="', 1) + 22)AS end_pos
    from   test
  )
);

这两者的输出结果如下:

WALLET_DIR
zurich_WALLET_DIR
英文:

You can use:

SELECT REGEXP_SUBSTR(
         HOST,
         '\(MY_WALLET_DIRECTORY\s*=\s*"(.*?)"\)',
         1,
         1,
         'i',
         1
       ) AS wallet_dir
from   test;

or, more to type but simple string functions are much faster than regular expressions:

SELECT CASE
       WHEN start_pos > 0 AND end_pos > 0
       THEN SUBSTR(host, start_pos + 22, end_pos - start_pos - 22)
       END AS wallet_dir
FROM   (
  SELECT host,
         INSTR(host, '(MY_WALLET_DIRECTORY="', 1) AS start_pos,
         INSTR(host, '")', INSTR(host, '(MY_WALLET_DIRECTORY="', 1) + 22)AS end_pos
  from   test
);

Which, for the sample data:

CREATE TABLE test (host) AS
SELECT '(DESCRIPTION=(ADDRESS=(PROTOCOL=tcps)(HOST="test1.zurich.com")(PORT=1521))(CONNECT_DATA=(SERVICE_NAME=apps_test1.zurich.com))(SECURITY=(MY_WALLET_DIRECTORY="/u01/zurich/Fdbb48380hdsvh8/data/zurich_WALLET_DIR")(SSL_SERVER_DN_MATCH=FALSE)))' FROM DUAL;

Both output:

WALLET_DIR
/u01/zurich/Fdbb48380hdsvh8/data/zurich_WALLET_DIR

If you just want the last part of the string then:

SELECT REGEXP_SUBSTR(
         HOST,
         '\(MY_WALLET_DIRECTORY\s*=\s*"[^"]*/(.*?)"\)',
         1,
         1,
         'i',
         1
       ) AS wallet_dir
from   test;

or:

SELECT SUBSTR(wallet_dir, INSTR(wallet_dir, '/', -1) + 1) AS wallet_dir
FROM   (
  SELECT CASE
         WHEN start_pos > 0 AND end_pos > 0
         THEN SUBSTR(host, start_pos + 22, end_pos - start_pos - 22)
         END AS wallet_dir
  FROM   (
    SELECT host,
           INSTR(host, '(MY_WALLET_DIRECTORY="', 1) AS start_pos,
           INSTR(host, '")', INSTR(host, '(MY_WALLET_DIRECTORY="', 1) + 22)AS end_pos
    from   test
  )
);

Which both output:

WALLET_DIR
zurich_WALLET_DIR

fiddle

答案2

得分: 1

尝试这个正则表达式:

select regexp_substr(HOST,'MY_WALLET_DIRECTORY=.*\/(.*)"',1,1,NULL,1) as Extracted
from mytable;

结果:

EXTRACTED
zurich_WALLET_DIR

演示在这里

英文:

Try this Regex :

select regexp_substr(HOST,'MY_WALLET_DIRECTORY=.*\/(.*)"\)',1,1,NULL,1) as Extracted
from mytable;

Result :

EXTRACTED
zurich_WALLET_DIR

Demo here

huangapple
  • 本文由 发表于 2023年3月9日 21:07:40
  • 转载请务必保留本文链接:https://go.coder-hub.com/75685046.html
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定