Linux – Remove all symbolic links of files in one command

This bash command should do it for you:

find /home/user/public_html/qa/ -type l -lname '/home/user/public_html/dev/*' -printf 'ln -nsf "$(readlink "%p" | sed s/dev/qa/)" "$(echo "%p" | sed s/dev/qa/)"\n' > script.sh

It uses find to identify all files in the qa directory that are symbolic links with a target that's in the dev directory, and for each one, it prints out a bash command that will replace the link with a link to the equivalent path in qa/. After you run this, just execute the generated script with

bash script.sh

You might want to examine it manually first to make sure it worked.

Here's a more verbose version of the above find command for easier reading (though I wouldn't necessarily write it this way in practice):

SRC_DIR="/home/user/public_html/qa"
OLD_TARGET="/home/user/public_html/dev"
SUB="s/dev/qa/"

find $SRC_DIR -type l \
  -lname "$OLD_TARGET/*" -printf \
  'ln -nsf "$(readlink "%p"|sed $SUB)" "$(echo "%p"|sed $SUB)"\n'\
 > script.sh

Not sure about UNIX cp implemebntation, but GNU cp by default is not preserving symlinks. And hard links are not the same thing as files. By default cp creates files/dirs from symlinks. Probably you don't need -L option for your purposes. If you really need this that you can use --no-preserve=links option.

Refer to man cp:

   --preserve[=ATTR_LIST]
          preserve the specified attributes (default: mode,ownership,timestamps), if possible additional attributes: context, links, xattr, all

   --no-preserve=ATTR_LIST
          don't preserve the specified attributes