Shell script for replacing some files with symlinks

ftpshellsvn

Some servers only support ftp to upload files.

When I export the a project from my subversion repository to my windows machine, all (linux) symlinks are replaces by placeholder-files:

link ../www_public/images

after uploading the all exported files I now use

find | xargs grep -P ^link

to find all those placeholders. I then replace them with the actual symlink manually.

I would really like to automate this step with a shell-script.
How would I do this?

Note:
If there is a better / different solution to this problem, don't hesitate to share it 🙂

Best Solution

Here's one possible solution:

:
grep -lr '^link' . | while read placeholderfile
do
  linkfile=`cut -c6- "$placeholderfile"`
  ln -sf "$linkfile" "$placeholderfile"
done

edit: changed code above w.r.t. the comments below.

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

To check if a directory exists in a shell script, you can use the following:

if [ -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY exists.
fi

Or to check if a directory doesn't exist:

if [ ! -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY doesn't exist.
fi

However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check. E.g. running this:

ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then 
  rmdir "$SYMLINK" 
fi

Will produce the error message:

rmdir: failed to remove `symlink': Not a directory

So symbolic links may have to be treated differently, if subsequent commands expect directories:

if [ -d "$LINK_OR_DIR" ]; then 
  if [ -L "$LINK_OR_DIR" ]; then
    # It is a symlink!
    # Symbolic link specific commands go here.
    rm "$LINK_OR_DIR"
  else
    # It's a directory!
    # Directory command goes here.
    rmdir "$LINK_OR_DIR"
  fi
fi

Take particular note of the double-quotes used to wrap the variables. The reason for this is explained by 8jean in another answer.

If the variables contain spaces or other unusual characters it will probably cause the script to fail.

Unix – Use grep –exclude/–include syntax to not grep through certain files

Use the shell globbing syntax:

grep pattern -r --include=\*.cpp --include=\*.h rootdir

The syntax for --exclude is identical.

Note that the star is escaped with a backslash to prevent it from being expanded by the shell (quoting it, such as --include="*.cpp", would work just as well). Otherwise, if you had any files in the current working directory that matched the pattern, the command line would expand to something like grep pattern -r --include=foo.cpp --include=bar.cpp rootdir, which would only search files named foo.cpp and bar.cpp, which is quite likely not what you wanted.

Update 2021-03-04

I've edited the original answer to remove the use of brace expansion, which is a feature provided by several shells such as Bash and zsh to simplify patterns like this; but note that brace expansion is not POSIX shell-compliant.

The original example was:

grep pattern -r --include=\*.{cpp,h} rootdir

to search through all .cpp and .h files rooted in the directory rootdir.

Best Solution

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

Unix – Use grep –exclude/–include syntax to not grep through certain files

Related Question