Rename multiple filename in multiple folders - bash

Question

I know you can do this to rename all filenames in a single folder with something like this:
for file in 1_*; do
mv "$file" "${file/1_/}"
done
However, is there a way to do this across multiple folders? For example, it will search through all the folders in the current directory and change them.
I have bash version 4.3

Answer 1

A robust solution, assuming you have GNU or BSD/OSX find:
find . -type f -name '1_*' -execdir sh -c 'echo mv -- "$1" "${1#1_}"' _ {} \;
Note:
- This will only echo the mv commands, to be safe; remove the echo to perform actual renaming.
- The OP's substitution, "${file/1_/}" was changed to the POSIX-compliant "${file#1_}", which is actually closer to the intent.
- If you truly need a substitution such as "${file/1_/}", which the sh on your system may or may not support, it is better to explicitly invoke a shell known to support it, such as bash.
- Symlinks are ignored (both files and directories); use find -L ... to include them (both as potential files to be renamed and to make find descend into symlinks to directories).
find . -type f -name '1_*' finds all files (-type f) with names matching 1_* (-name '1_*') in the current dir.'s (.) subtree.
-execdir executes the command passed to it in the subdirectory in which the file at hand is located.
sh -c 'echo mv -- "$1" "${1#1_}"' _ {} \; invokes the default shell (sh):
with a command string (passed to -c)
mv -- "$1" "${1#1_}" effectively removes prefix 1_ from the filename represented by the first positional parameter ($1).
and dummy parameter _ (which sh will assign to $0, which is not of interest here)
and the path of the file at hand, {}, which the shell will bind to $1;
\; simply terminates -execdir's argument.
Note that -- ensures that any filename that happens to start with - isn't mistaken for an option by mv (applies analogously below).
-execdir is not POSIX-compliant; if a POSIX-compliant variant is therefore more cumbersome:
find . -type f -name '1_*' -exec sh -c \
'cd -- "${1%/*}" && f=${1##*/} && echo mv -- "$f" "${f#1_}"' _ {} \;
cd -- "${1%/*}" changes to the directory in which the file at hand is located.
Note: cd -- "$(dirname -- "$1")" is generally more robust, but less efficient; since we know that $1 is always a path rather than a mere filename in this scenario, we can use the more efficient cd -- "${1%/*}".
f=${1##*/} extracts the mere filename from the file path at hand.
The remainder of the command line then works as above, analogously.
Performance note:
The above solutions are concise, but inefficient, because a shell instance has to be spawned for each matching file.
A potential speed-up is to use a variant of peak's approach, but only if you avoid calling external utilities in the while loop (except for mv):
find . -type f -name '1_*' | while IFS= read -r f; do
new_name=${f##*/} # extract filename
new_name=${new_name#1_} # perform substitution
d=${f%/*} # extract dir
echo mv -- "$f" "$d/$new_name" # call mv to rename
done
The above bears the hypothetical risk of breaking with filenames with embedded newlines (very rare); with GNU or BSD find, this problem could be solved.
With this approach, only a single shell instance is spawned, which processes all matching filenames in a loop - as long as the only external utility that is called in the loop is mv, this will generally be faster than the find-only solutions with -exec[dir].

Answer 2

If you don't want to depend on too many subtleties, consider this very pedestrian approach, which assumes a bash-like shell and that all the usual suspects (find, sed, ....) are directly available:
find . -type f -name "1_*" | while read -r file ; do
x=$(basename "$file")
y=$(sed 's/1_//' <<< "$x")
d=$(dirname "$file")
mv "$file" "$d/$y"
done
(You might want to try this using "mv -i" or "echo mv ....". You might also want to use find with the -follow option.)