regex在特定字符后匹配或不匹配

I want to match the rule with or without redundant string in the end. And swap the match patterns.
Example:

我想要匹配规则是否有冗余字符串。交换匹配模式。例子:

$　"aaa bbb ccc"　--> "bbb aaa"
$ "aaa bbb" --> "bbb aaa"

Here is what I tried but does not work.

这是我尝试过但没有成功的方法。

echo "aaa bbb ccc" | sed -e 's:\(.*\)\s\(.*\)(\s\(.*\))?:\2\s\1:g'
$ aaa bbb ccc

Thank you.

谢谢你！

2 个解决方案

#1

Using sed

Try:

试一试:

$ echo "aaa bbb" | sed -Ee 's:([[:alnum:]]+)[[:space:]]+([[:alnum:]]+).*:\2 \1:'
bbb aaa
$ echo "aaa bbb ccc" | sed -Ee 's:([[:alnum:]]+)[[:space:]]+([[:alnum:]]+).*:\2 \1:'
bbb aaa

Notes:

注:

1. sed recognizes POSIX regular expressions. \s is a GNU-only extension. Character classes like [[:space:]] and [[:alnum:]] are understand by all implementations of sed.

   sed识别POSIX正则表达式。\s是gnuonly的扩展。像[:space:]和[[:alnum:]这样的字符类被sed的所有实现所理解。

2. I added the option -E to get extended regular expressions. That eliminates the need for the backslashes and it also enables to the use of + to mean one or more.

   我添加了选项-E来扩展正则表达式。这就消除了对反斜杠的需要，并且还允许使用+来表示一个或多个。

3. There is no need to match parts of the line after the words that you want to exchange.

   不需要在你想要交换的词之后匹配行中的部分。

Using bash

$ read a b c<<<"aaa bbb"; echo "$b $a"
bbb aaa
$ read a b c<<<"aaa bbb ccc"; echo "$b $a"
bbb aaa

Using awk

$ echo "aaa bbb" | awk '{print $2,$1}'
bbb aaa
$ echo "aaa bbb ccc" | awk '{print $2,$1}'
bbb aaa

#2

awk solution:

awk的解决方案:

echo aaa bbb | awk '{print $2" "$1}'
echo aaa bbb ccc | awk '{print $2" "$1}'

what it does is just print the 1st and the 2nd fields in the reverse order ignoring the rest of the lines: elements are identified by the default FS of awk that can be adapted to your needs

它所做的只是以相反的顺序打印第1和第2个字段，而忽略其余行:元素由awk的默认FS标识，可以根据您的需要进行调整

Hope it helps you!

希望它能帮助你!