Linux sed - 如何使用 sed 做正则表达式组
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sed - how to do regex groups using sed
提问by Shengjie
Is there anyway you can do regex match group using sed like java regex pattern/match/group?
无论如何你可以像java regex pattern/match/group一样使用sed来做regex match group吗？
if i have string like
如果我有像这样的字符串
test-artifact-201251-balbal-0.1-SNAPSHOT.jar
how do I use sed just to get the result like:
我如何使用 sed 来获得如下结果：
test-artifact-0.1-SNASHOT.jar
I am wondering does sed allow you to do something like java regex, you define the pattern like:
我想知道 sed 是否允许您执行类似 java regex 的操作，您定义的模式如下：
([a-z]*-[a-z]*-)([0-9]*-)([a-z]*-)([.]*SNAPSHOT.jar)
and then you can get the results as an array like:
然后你可以得到一个数组的结果，如：
test-artifact-
201251-
balbal-
0.1-SNAPSHOT.jar
采纳答案by Birei
You have to escape parentheses to group expressions:
您必须转义括号才能对表达式进行分组：
\([a-z]*-[a-z]*-\)\([0-9]*-\)\([a-z]*-\)\([.]*SNAPSHOT.jar\)
And use them with \1, \2, etc.
并与使用它们\1，\2等等。
EDIT: Also note just before SNAPSHOTthat [.]will not match. Inside brackets .is literal. It should be [0-9.-]*
编辑：还要注意之前SNAPSHOT那[.]将不匹配。括号内.是字面意思。它应该是[0-9.-]*
回答by Steve
This is what Birei and Thor mean:
这就是 Birei 和 Thor 的意思：
sed -r "s/([a-z]*-[a-z]*-)([0-9]*-)([a-z]*-)(.*)/\n\n\n/"
Output:
输出：
test-artifact-
201251-
balbal-
0.1-SNAPSHOT.jar
回答by Kent
infact for those regular string, awk could save you from grouping. :)
事实上，对于那些常规字符串，awk 可以使您免于分组。:)
you just give the part index number you want:
您只需提供所需的零件索引号：
awk 'BEGIN{FS=OFS="-"}{print ,,,}'
output:
输出：
kent$ echo "test-artifact-201251-balbal-0.1-SNAPSHOT.jar"|awk 'BEGIN{FS="-";OFS="-"}{print ,,,}'
test-artifact-0.1-SNAPSHOT.jar
回答by SAM80DEV
If you are searching for an easier way I guess this might be of your help! :)
如果您正在寻找一种更简单的方法，我想这可能对您有所帮助！:)
echo "est-artifact-201251-balbal-0.1-SNAPSHOT.jar" | cut -d- -f1,2,5,6
"-" used as delimeter and fields 1,2,5,6 are printed.
“-”用作分隔符并打印字段 1,2,5,6。
Note: This would require you to know the exact position of the field.
注意：这需要您知道该字段的确切位置。
