linux find more than one -exec - bash

find . -iname "*.txt" -exec program '{}' \; | sed 's/Value= //'
-"program" returns a different value for each file, and the output is prefixed with "Value= "
In this time the output will be "Value= 128" and the after sed just 128.
How can I take just the value "128" and have the input file be renamed to 128.txt
but also have this find run thought multiple files.
sorry for bad descriptions.
I will try to clear if needed

First write a shell script capable of renaming an argument:
mv "$1" "$(program "$1" | sed "s/Value= //").txt"
Then embed that script in your find command:
find . -iname "*.txt" \
-exec sh -c 'mv "$1" "$(program "$1" | sed "s/Value= //").txt"' _ {} \;

Related

Solution for find -exec if single and double quotes already in use

I would like to recursively go through all subdirectories and remove the oldest two PDFs in each subfolder named "bak":
Works:
find . -type d -name "bak" \
-exec bash -c "cd '{}' && pwd" \;
Does not work, as the double quotes are already in use:
find . -type d -name "bak" \
-exec bash -c "cd '{}' && rm "$(ls -t *.pdf | tail -2)"" \;
Any solution to the double quote conundrum?
In a double quoted string you can use backslashes to escape other double quotes, e.g.
find ... "rm \"\$(...)\""
If that is too convoluted use variables:
cmd='$(...)'
find ... "rm $cmd"
However, I think your find -exec has more problems than that.
Using {} inside the command string "cd '{}' ..." is risky. If there is a ' inside the file name things will break and might execcute unexpected commands.
$() will be expanded by bash before find even runs. So ls -t *.pdf | tail -2 will only be executed once in the top directory . instead of once for each found directory. rm will (try to) delete the same file for each found directory.
rm "$(ls -t *.pdf | tail -2)" will not work if ls lists more than one file. Because of the quotes both files would be listed in one argument. Therefore, rm would try to delete one file with the name first.pdf\nsecond.pdf.
I'd suggest
cmd='cd "$1" && ls -t *.pdf | tail -n2 | sed "s/./\\\\&/g" | xargs rm'
find . -type d -name bak -exec bash -c "$cmd" -- {} \;
You have a more fundamental problem; because you are using the weaker double quotes around the entire script, the $(...) command substitution will be interpreted by the shell which parses the find command, not by the bash shell you are starting, which will only receive a static string containing the result from the command substitution.
If you switch to single quotes around the script, you get most of it right; but that would still fail if the file name you find contains a double quote (just like your attempt would fail for file names with single quotes). The proper fix is to pass the matching files as command-line arguments to the bash subprocess.
But a better fix still is to use -execdir so that you don't have to pass the directory name to the subshell at all:
find . -type d -name "bak" \
-execdir bash -c 'ls -t *.pdf | tail -2 | xargs -r rm' \;
This could stll fail in funny ways because you are parsing ls which is inherently buggy.
You are explicitely asking for find -exec. Usually I would just concatenate find -exec find -delete but in your case only two files should be deleted. Therefore the only method is running subshell. Socowi already gave nice solution, however if your file names do not contain tabulator or newlines, another workaround is find while read loop.
This will sort files by mtime
find . -type d -iname 'bak' | \
while read -r dir;
do
find "$dir" -maxdepth 1 -type f -iname '*.pdf' -printf "%T+\t%p\n" | \
sort | head -n2 | \
cut -f2- | \
while read -r file;
do
rm "$file";
done;
done;
The above find while read loop as "one-liner"
find . -type d -iname 'bak' | while read -r dir; do find "$dir" -maxdepth 1 -type f -iname '*.pdf' -printf "%T+\t%p\n" | sort | head -n2 | cut -f2- | while read -r file; do rm "$file"; done; done;
find while read loop can also handle NUL terminated file names. However head can not handle this, so I did improve other answers and made it work with nontrivial file names (only GNU + bash)
replace 'realpath' with rm
#!/bin/bash
rm_old () {
find "$1" -maxdepth 1 -type f -iname \*.$2 -printf "%T+\t%p\0" | sort -z | sed -zn 's,\S*\t\(.*\),\1,p' | grep -zim$3 \.$2$ | xargs -0r realpath
}
export -f rm_old
find -type d -iname bak -execdir bash -c 'rm_old "{}" pdf 2' \;
However bash -c might still exploitable, to make it more secure let stat %N do the quoting
#!/bin/bash
rm_old () {
local dir="$1"
# we don't like eval
# eval "dir=$dir"
# this works like eval
dir="${dir#?}"
dir="${dir%?}"
dir="${dir//"'$'\t''"/$'\011'}"
dir="${dir//"'$'\n''"/$'\012'}"
dir="${dir//$'\047'\\$'\047'$'\047'/$'\047'}"
find "$dir" -maxdepth 1 -type f -iname \*.$2 -printf '%T+\t%p\0' | sort -z | sed -zn 's,\S*\t\(.*\),\1,p' | grep -zim$3 \.$2$ | xargs -0r realpath
}
find -type d -iname bak -exec stat -c'%N' {} + | while read -r dir; do rm_old "$dir" pdf 2; done

Printing the shell find and remove command to screen and log file

I have a script that finds log files older than x days within a specified directory and removes them.
find $LOG_ARCHIVE/* -mtime +$DAYS_TO_KEEP_LOGS -exec rm -f {} \;
This is working as expected but I would like to have the option to print the processing to the screen and log file so I know what files (if any) have been deleted. I've tried appending tee at the end but have had no success.
find $LOG_ARCHIVE/* -mtime +$DAYS_TO_KEEP_LOGS -exec rm -fv {} \; | tee -a $LOG
There are multiple ways the task can be done.
One possibility is to simply run find twice:
find "$LOG_ARCHIVE" -mtime +"$DAYS_TO_KEEP_LOGS" -print > "$LOG"
find "$LOG_ARCHIVE" -mtime +"$DAYS_TO_KEEP_LOGS" -exec rm -f {} +
Another possibility is to use tee along with (GNU extensions) -print0 to find and -0 to xargs:
find "$LOG_ARCHIVE" -mtime +"$DAYS_TO_KEEP_LOGS" -print0 |
tee "$LOG" |
xargs -0 rm -f
With this version, the log file will have null bytes at the end of each file name. You can arrange to replace those with newlines if you don't mind the possible ambiguity:
find "$LOG_ARCHIVE" -mtime +"$DAYS_TO_KEEP_LOGS" -print0 |
tee >(tr '\0' '\n' >"$LOG") |
xargs -0 rm -f
This uses Bash (and Korn shell) process substitution to pass the log file through tr to map null bytes '\0' to newlines '\n'.
Another way of doing it is to write a tiny custom script (call it remove-log.sh):
printf '%s\n' "$#" >> "$LOG"
rm -f "$#"
and then use:
find "$LOG_ARCHIVE" -mtime +"$DAYS_TO_KEEP_LOGS" -exec bash remove-log.sh {} +
Note that the script needs to see the value of $LOG, so that must be exported as an environment variable. You could avoid that by passing the log name explicitly:
logfile="$1"
shift
printf '%s\n' "$#" >> "$logfile"
rm -f "$#"
plus:
find "$LOG_ARCHIVE" -mtime +"$DAYS_TO_KEEP_LOGS" -exec bash remove-log.sh "$LOG" {} +
Note that both of these use >> to append because the script might be invoked more than once (though it probably won't be). The onus is on you to ensure that the log file is empty before you run the find command.
Note that I dropped the /* from the path argument for find; it wasn't really needed. You might want to add -type f to ensure that only files are removed. The + is a feature from the POSIX 2008 specification of find which makes find act rather like xargs without needing to explicitly use xargs.
find $LOG_ARCHIVE/* -mtime +$DAYS_TO_KEEP_LOGS -exec sh -c 'echo {} |tee -a "$LOG"; rm -f {}' \;
Try and see if it works.

issue with piping find into sed (find and replace)

Here is my current code, my goal is to find every file in a given directory (recursively) and replace "FIND" with "REPLACEWITH" and overwrite the files.
FIND='ALEX'
REPLACEWITH='<strong>ALEX</strong>'
DIRECTORY='/some/directory/'
find $DIRECTORY -type f -name "*.html" -print0 |
LANG=C xargs -0 sed -i "s|$FIND|$REPLACEWITH|g"
The error I am getting is:
sed: 1: "/some/directory ...": command a expects \ followed by text
As given in BashFAQ #21, you can use perl to perform search-and-replace operations with no potential for data being treated as code:
in="$FIND" out="$REPLACEWITH" find "$DIRECTORY" -type f -name '*.html' \
-exec perl -pi -e 's/\Q$ENV{"in"}/$ENV{"out"}/g' '{}' +
If you want to include only files matching the FIND string, find can be told to only pass files which grep flags on to perl:
in="$FIND" out="$REPLACEWITH" find "$DIRECTORY" -type f -name '*.html' \
-exec grep -F -q -e "$FIND" '{}' ';' \
-exec perl -pi -e 's/\Q$ENV{"in"}/$ENV{"out"}/g' '{}' +
Because grep is being used to evaluate individual files, it's necessary to use one grep call per file so its exit status can be evaluated on a per-file basis; thus, the use of the less efficient -exec ... {} ';' action. For perl, it's possible to put multiple files to process on one command, hence the use of -exec ... {} +.
Note that fgrep is line-oriented; if your FIND string contains multiple lines, then files with any one of those lines will be passed to perl for replacements.
You can have find invoke sed directly although I think all the modification times on your files will be affected (which might matter or not):
find $DIRECTORY -type f -name "*.html" -exec sed -i "s|$FIND|$REPLACEWITH|g" '{}' ';'

Replacing a part of file path in exec

I would like to replace the part of each file path, which will be find by find linux command.
My approach is attached below:
find . -type f -name "*.txt" -exec echo {} | sed "s/f/u/g" {} \;
I expect the replacement of each letter "f" with "u" in file path. Unfortunately I got this error:
find: missing argument to `-exec'
sed: can't read {}: No such file or directory
sed: can't read ;: No such file or directory
What I did wrong? Thank you for your help.
I would like to replace the part of each file path
If you want to change just the file names/paths then use:
find . -type f -name "*.txt" -exec bash -c 'echo "$1" | sed "s/f/u/g"' - {} \;
or a bit more efficient with xargs (since it avoids spawning subshell for each found file):
find . -type f -name "*.txt" -print0 |
xargs -0 bash -c 'for f; do sed "s/f/u/g" <<< "$f"; done'
find . -type f -name "*.txt" | while read files
do
newname=$(echo "${files}" | sed s"#f#u#"g)
mv -v "${files}" "${newname}"
done
I don't completely understand what you meant by file path. If you weren't talking about the file name, please clarify further.

Bash script to change file names recursively

I have a script for changing file names of mht files but it does not traverse through dirs and sub dirs. I asked a question on a local forum and I got an answer that this is a solution:
find . -type f -name "*.mhtml" -o -type f -name "*.mht" | xargs -I item sh -c '{ echo item; echo item | sed "s/[:?|]//g"; }' | xargs -n2 mv
But it generates an error. With some of my experimenting it turns out that sh -c breaks file names with space and that this generates an error. How can I fix this?
#!/bin/bash
# renames.sh
# basic file renamer
for i in . *.mht
do
j=`echo $i | sed 's/|/ /g' | sed 's/:/ /g' | sed 's/?//g' | sed 's/"//g'`
mv "$i" "$j"
done
#! /bin/bash
find . -type f \( -name "*.mhtml" -o -name ".mht" \) -print0 |
while IFS= read -r -d '' source; do
target="${source//[:?|]/}"
[ "X$source" != "X$target" ] &&
mv -nv "$source" "$target"
done
Update: Do the rename according to the original question, and added support for .mht.
Use rename. With rename you can specify a renaming pattern:
find . -type f \( -name "*.mhtml" -o -name "*.mht" \) -print0 | xargs -0 -I'{}' rename 's/[:?|]//g' "{}"
This way you can properly handle names with spaces. xargs will replace {} with every names of file provided by the find command. Also note the use of -print0 and -0. This use a \0 as a separator so its avoid problems dealing with filnames containing \n (newline).
The -o was not working the way it was intended to. you must use parenthesis to group conditions.
You may also consider using -iname instead of -name if you deal with file ending with ".mHtml".

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