如何使用grep在`name =`之后获取任何内容? [英] How to use grep to get anything just after `name=`?

问题描述

我一直在尝试 grep name= 之后的任何内容，只包含空格和字母数字.

I’m stuck in trying to grep anything just after name=, include only spaces and alphanumeric.

例如:

name=some value here

我明白

some value here

我完全是新手，下面的 grep 匹配包括 name= 在内的所有内容.

I’m totally newb in this, the following grep match everything including the name=.

grep 'name=.*' filename

非常感谢任何帮助.

推荐答案

详细这里，你想要一个肯定的lookbehind子句，比如:

As detailed here, you want a positive lookbehind clause, such as:

grep -P '(?<=name=)[ A-Za-z0-9]*' filename

-P 使 grep 使用 Perl 方言，否则您可能需要转义括号.您还可以如其他地方所述，附加 -o 参数以仅打印出匹配的内容.括号中的部分指定您需要字母数字和空格.

The -P makes grep use the Perl dialect, otherwise you'd probably need to escape the parentheses. You can also, as noted elsewhere, append the -o parameter to print out only what is matched. The part in brackets specifies that you want alphanumerics and spaces.

使用肯定的lookbehind 子句的优点是name="文本不是匹配的一部分.如果 grep 突出显示匹配的文本，它只会突出显示字母数字(和空格)部分.-o 参数也不会显示name="部分.而且，如果您将其转换为另一个程序，例如 sed，它可能会捕获文本并对其进行处理，您将不会捕获name="部分，尽管您也可以通过使用捕获括号来实现这一点.

The advantage of using a positive lookbehind clause is that the "name=" text is not part of the match. If grep highlights matched text, it will only highlight the alphanumeric (and spaces) part. The -o parameter will also not display the "name=" part. And, if you transition this to another program like sed that might capture the text and do something with it, you won't be capturing the "name=" part, although you can also do that by using capturing parenthess.

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