本文介绍了在Postgres中查找和替换正则表达式的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个表,该表包含许多行,其中的列包含一个URL。 URL的格式为:
I have a table that contains a number of rows with columns containing a URL. The URL is of the form:
one.example1:9999/dotFile
我想将该列中的所有匹配项替换为 example2/dotFile 同时保留:9999之后的所有内容。我在regexp_matches和regexp_replace上找到了一些文档,但是我不太了解。
I would like to replace all matches in that column with example2/dotFile while retaining everything after :9999. I have found some documentation on regexp_matches and regexp_replace, but I can't quite wrap my head around it.
推荐答案如果您知道网址,则不必使用正则表达式。 replace()函数适合您:
if you know the url, you don't have to use regex. replace() function should work for you:
replace(string text, from text, to text) Replace all occurrences in string of substring from with substring to example: replace('abcdefabcdef', 'cd', 'XX') abXXefabXXef您可以尝试:
replace(yourcolumn, 'one.example1:9999','example2')更多推荐
在Postgres中查找和替换正则表达式
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