linux - 使用 sed 显示文件的确切位置

Question

我的命令是:

[root@my /]# grep -lr --include=*version.php "$wp_version"/home/draka/www/
/home/draka/www/wp-content/version.php
/home/draka/www/wp-content/themes/version.php
/home/draka/www/wp-includes/version.php

然后我使用 sed 来查找特定行:

[root@my /]# grep -lr --include=*version.php "$wp_version" /home/draka/www/ | xargs sed -n '7p'
$wp_version = '3.5';

每个文件夹的文件版本不同:

1. /home/draka/www/wp-content/version.php = $wp_version = '3.5.1';
2. /home/draka/www/wp-content/themes/version.php = $wp_version = '3.5';
3. /home/draka/www/wp-includes/version.php = $wp_version = '1.5';

我的问题是，如何合并这 2 个输出？我想要的输出可能是这样的:

/home/draka/www/wp-content/version.php = $wp_version = '3.5.1';

或

/home/draka/www/wp-content/version.php -> $wp_version = '3.5.1';

非常感谢您的及时回复。谢谢。

Answer 1

省略 -l，比如 grep -r --include=*version.php wp_version/home/draka/www/，就可以了。另见 man grep:

-H, --with-filename
   Print the file name for each match.  This is the default when there is more
   than one file to search.