我想在字段中获取电子邮件格式的文本。我在下面尝试了sql，但没有运气。见 SqlFiddle .从正则表达式中删除 ^ 和 $ 也不起作用。
WITH TEST_DATA AS (
SELECT 'foo@gmail.com' AS EMAIL FROM DUAL UNION ALL
SELECT 'mail foo@gmail.com' FROM DUAL UNION ALL
SELECT 'mail foo@gmail.com sent' FROM DUAL UNION ALL
SELECT 'foo@gmail.com sent count 23' FROM DUAL UNION ALL
SELECT 'mail already sent to foo@gmail.com and foo@hotmail.com' FROM DUAL UNION ALL
SELECT 'foo@gmail.com sent count 23' FROM DUAL
)SELECT REGEXP_SUBSTR(EMAIL,'^[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}$') MAIL
FROM TEST_DATA;
此数据集的预期输出
foo@gmail.com
foo@gmail.com
foo@gmail.com
foo@gmail.com
foo@gmail.com, foo@hotmail.com
foo@gmail.com
任何帮助表示赞赏。
最佳答案
如果要在单个列中提取多个邮件ID，可以使用REGEXP_REPLACE函数。
假设您数据中的所有 id 都是有效的，
REGEXP_REPLACE (EMAIL, '(\w+@\w+\.\w+ ?)|(.)', '\1')
这将删除所有其他文本，除了至少由空格分隔的邮件 ID。
然后您可以删除任何尾随空格并添加逗号以分隔多个 ID。
REPLACE (TRIM (REGEXP_REPLACE (EMAIL, '(\w+@\w+\.\w+ ?)|(.)', '\1')),
' ',
', ')
例子:
WITH TEST_DATA
AS (SELECT 'foo@gmail.com' AS EMAIL FROM DUAL
UNION ALL
SELECT 'mail foo@gmail.com' FROM DUAL
UNION ALL
SELECT 'mail foo@gmail.com sent to 123@zxc.com and qwe@rt.com' FROM DUAL
UNION ALL
SELECT 'foo@gmail.com sent count 23 and asd@qwert.edu' FROM DUAL
UNION ALL
SELECT 'mail already sent to foo@gmail.com and foo@hotmail.com' FROM DUAL
UNION ALL
SELECT 'foo@gmail.com sent count 23' FROM DUAL)
SELECT REPLACE (TRIM (REGEXP_REPLACE (EMAIL, '(\w+@\w+\.\w+ ?)|(.)', '\1')),
' ',
', ')
MAIL
FROM TEST_DATA;
MAIL
-----------------------------
foo@gmail.com
foo@gmail.com
foo@gmail.com, 123@zxc.com, qwe@rt.com
foo@gmail.com, asd@qwert.edu
foo@gmail.com, foo@hotmail.com
foo@gmail.com
关于sql - 使用 Oracle Regexp 从字段中提取电子邮件，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21286245/