我的命令是:
[root@my /]# grep -lr --include=*version.php "$wp_version"/home/draka/www/
/home/draka/www/wp-content/version.php
/home/draka/www/wp-content/themes/version.php
/home/draka/www/wp-includes/version.php
我用sed找到具体的行:
[root@my /]# grep -lr --include=*version.php "$wp_version" /home/draka/www/ | xargs sed -n '7p'
$wp_version = '3.5';
每个文件夹的文件版本不同:
/主页/draka/www/wp content/version.php=$wp_version='3.5.1';
/home/draka/www/wp content/themes/version.php=$wp_version='3.5';
/home/draka/www/wp includes/version.php=$wp_version='1.5';
我的问题是,我如何将这两个输出结合起来?我想要的输出可能是:
/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';
非常感谢你的及时答复。谢谢您。
最佳答案
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.
关于linux - 使用sed显示文件的确切位置,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/16359410/