sed替换和外部命令 [英] Sed substitution and external command

问题描述

我有一个这种格式的输入日志文件

I have an input log file in this format

May 23 2012 11:59:56
a;b;c
May 21 2012 16:54:12
d;e;f
May 19 2012 16:22:52
g;h;i
...

我想以这种格式输出

2012-05-23
a;b;c
2012-05-21
d;e;f
2012-05-19
g;h;i
...

使用 sed ，我知道如何替换日期行

Using sed, I know how to substitute the date lines

% sed 's/.*:.*:.*/match_string/' input.txt
match_string
a;b;c
match_string
d;e;f
match_string
g;h;i
...

使用 date ，我知道如何转换日期:

Using date, I know how to convert dates :

% date -d 'May 23 2012 11:59:56' '+%Y-%m-%d'
2012-05-23

但是在sed命令期间如何使match_string求值?

But how can make match_string to be evaluated during the sed command ?

推荐答案

如果有 GNU Sed 可用，则可以使用 e 标志:

If you have GNU Sed available, you could use the e flag:

sed 's/.*:.*/date -d"&" "+%Y-%m-%d"/ge' file

将为您的示例提供帮助，请参见测试:

will help you for your example, see the test:

kent$  echo "May 23 2012 11:59:56
a;b;c
May 21 2012 16:54:12
d;e;f
May 19 2012 16:22:52
g;h;i"|sed 's/.*:.*/date -d"&" "+%Y-%m-%d"/ge'
2012-05-23
a;b;c
2012-05-21
d;e;f
2012-05-19
g;h;i

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