bash脚本定期防爆pressions ...如何查找和替换所有的匹配？ [英] Bash Script Regular Expressions...How to find and replace all matches?

问题描述

我写一个bash脚本，逐行读取一个文件行。

I am writing a bash script that reads a file line by line.

该文件是一个包含格式DD / MM / YYYY许多日期的.csv文件，但我想将其更改为YYYY-MM-DD。

The file is a .csv file which contains many dates in the format DD/MM/YYYY but I would like to change them to YYYY-MM-DD.

我会使用常规的前pression相匹配的数据，并更换它使得所有文件中的日期正确格式为YYYY-MM-DD。

I would to match the data using a regular expression, and replace it such that all of the dates in the file are correctly formatted as YYYY-MM-DD.

我相信这个普通的前pression将匹配日期：

I believe this regular expression would match the dates:

([0-9][0-9]?)/([0-9][0-9]?)/([0-9][0-9][0-9][0-9])

但我不知道如何找到正则表达式匹配，并使用新的格式取代他们，或者如果这甚至有可能在bash脚本。请帮助！

But I do not know how to find regex matches and replace them with the new format, or if this is even possible in a bash script. Please help!

推荐答案

试试这个使用SED：

line='Today is 10/12/2010 and yesterday was 9/11/2010'
echo "$line" | sed -r 's#([0-9]{1,2})/([0-9]{1,2})/([0-9]{4})#\3-\2-\1#g'

OUTPUT:
Today is 2010-12-10 and yesterday was 2010-11-9

PS：在Mac上使用 SED -E 而不是 SED -r

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