从字符串oracle提取日期 [英] Extract date from string oracle
问题描述
我有一列中的字符串以-'伊利诺伊州芝加哥,2015年4月20日—等等这里的文字.我想从Oracle中的此字符串中提取Date部分.关于如何执行此操作的任何想法.我可以为mm/dd/yyyy找到以下内容,但不能找到长日期格式.
I have a column in which a string starts with - 'Chicago, IL, April 20, 2015 — and so on text here'. I want to extract the Date part from this string in Oracle. Any ideas on how to do this. I was able to find something for mm/dd/yyyy like below, but not for long date format.
SELECT REGEXP_SUBSTR(' the meeting will be on 8/8/2008', '[0-9]{1,}/[0-9]{1,}/[0-9]{2,}') FROM dual
推荐答案
如果您的列值始终以'Chicago, IL, April 20, 2015 — and so on text here'
开头,则可以简单地使用SUBSTR
而不是REGEXP_SUBSTR
If your columns value is always start with 'Chicago, IL, April 20, 2015 — and so on text here'
then you could simly use SUBSTR
instead of REGEXP_SUBSTR
SELECT
SUBSTR(column_name
,INSTR(column_name, ',', 1, 2) + 1
,INSTR(column_name, '—') - INSTR(column_name, ',', 1, 2) - 1
)
FROM
dual;
否则,您可以使用REGEXP_SUBSTR
作为其他答案,我原来的答案是错误的@MTO
评论
If not then you could use REGEXP_SUBSTR
as other answer mention, my original answer is wrong as @MTO
comment
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