问题描述
给出此文件:
a=as/dsdf b=fdfsf c=vcv
c=15 b=1 a=azzzz))]ee
a=12 z=19 r=15
我只想检索以 a =
因此在这种情况下的输出将是:
so the output in this case would be:
a=as/dsdf
a=azzzz))]ee
a=12
我已经深入研究bash文档，但找不到任何简单的方法，您有什么建议吗?
I've dived into bash documentation but couldn't find anything easy, do you have any suggestion ?
谢谢
推荐答案
我假设您希望以 a = 开头的所有内容都扩展到下一个空格.
I'm assuming that you want everything starting with a= up to the next space.
并非所有版本的grep都支持，但是如果您有 -o 选项，这很简单:
It's not supported by all versions of grep but if you have the -o option this is easy:
grep -Eo 'a=[^ ]+' file
-o 打印该行的匹配部分. -E 启用 扩展>正则表达式，使您可以使用 + 来表示一个或多个前一个 atom *的出现. [^] 表示空格以外的任何字符.
-o prints the matching part of the line. -E enables Extended Regular Expressions, which enables you to use + to mean one or more occurrences of the preceding atom*. [^ ] means any character other than a space.
否则，请使用sed捕获您感兴趣的零件:
Otherwise, use sed to capture the part you're interested in:
sed -E 's/.*(a=[^ ]+).*/\1/' file
作为万不得已的方法，如果您的sed版本不支持扩展的正则表达式，则该版本应适用于任何版本:
As a last resort, if your version of sed doesn't support extended regexes, this should work on any version:
sed 's/.*\(a=[^ ]\{1,\}\).*/\1/' file
正如注释中正确指出的(谢谢)一样，为避免打印不匹配的行，您可能还想使用 -n 来抑制输出并添加 p 命令以打印匹配的行:
As rightly pointed out in the comments (thanks), to avoid printing lines that don't match, you may also want to use -n to suppress output and add a p command to print lines that match:
sed -nE 's/.*(a=[^ ]+).*/\1/p' file
atom *:字符类，标记的子表达式或单个字符.
atom* : A character class, a marked sub-expression or a single character.
这篇关于在bash中，如何检索以特定表达式开头的字符串?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！