Need help on writing a bash script that will rename files that are being outputted as file name.suffix.date I need these files to be rewritten as name.date.suffix instead.
Edited:
Changed suffix from date to ~
Here's what I have so far:
find . -type f -name "*.~" -print0 | while read -d $'\0' f
do
new=`echo "$f" | sed -e "s/~//"`
mv "$f" "$new"
done
This changes the suffix back to original but can't figure out how to get the date to be named before the extension (fname??)
You can use regular expression matching to pull apart the original file name:
find . -type f -name "*.~" -print0 | while read -d $'\0' f
do
dir=${f%/*}
fname=${f##*/}
[[ $fname =~ (.+)\.([^.]+)\.([^.]+)\.~$ ]] || continue
name=${BASH_REMATCH[1]}
suffix=${BASH_REMATCH[2]}
d=${BASH_REMATCH[3]}
mv "$f" "$dir/$name.$d.$suffix"
done
Bash-only solution:
while IFS=. read -r -u 9 -d '' name suffix date tilde
do
mv "${name}.${suffix}.${date}.~" "${name}.${date}.${suffix}"
done 9< <(find . -type f -name "*.~" -print0)
Notes:
-d '' gives you the same result as -d $'\0'
Splits file names by the dots while reading them. Of course this means it would break if there are dots anywhere else.
Should otherwise work with pretty much any filenames, including those containing space, newlines and other funny business.
create a list of the files first and redirect to a file.
ls > fileList.txt
Open the file and read line by line in Perl. Use a regex to match the parts of the files and capture them like this
my ($fileName,$suffix,$date)=($WholeFileName=~/(.*)\.(.*)\.(.*)/);
This should capture the three seperate variables for you. Now all you need to do is move the old file to the new file name. The new file name will be a concatenation of the above three variables that you have got. $newFileName=$fileName. ".".$date.".".$suffix. If you have a sample fileName post a comment and I can reply with a short script. Perl is not the only way. You could just use bash or awk and find alternate ways to do this.
cut each part of your filenames:
FIN=$(echo test.12345.ABCDEF | sed -e 's/[a-zA-Z0-9]*[\\.][a-zA-Z0-9]*[\\.]//')
DEBUT=$(echo test.12345.ABCDEF | sed -e 's/[\\.][a-zA-Z0-9]*[\\.][a-zA-Z0-9]*//')
MILIEU=$(echo test.12345.ABCDEF | sed -e 's/'${FIN}'//' -e 's/'${DEBUT}'//' -e 's/[\.]*//g')
paste each part as expected:
echo ${DEBUT}.${FIN}.${MILIEU}
rename --no-act 's/\(name-regex\).\(suffix-regex\).\(date-regex\)/\1.\3.\2' *
Tweak the three regexes to fit your file names, and remove --no-act when you're happy with the result to actually rename the files.
Related
All:
Quickly and succinctly, I have many many files named as such:
lorem(12312315).txt
ipsum(578938-12-315-13-416-4).txt
amet(ran-dom-guid).txt
And I want to rename them to what's inside the parentheses dot text, like so:
12312315.txt
578938-12-315-13-416-4.txt
randomguid.txt
I'm sure a mix of sed, awk, grep, etc will do it, but commenting out the parentheses from the shell is throwing me. I cant come up with a string that will do it.
If anyone is kind enough to share a few thought cycles and help me, it would be a lovely Karma gesture!
Thanks for reading!
-Jim
Another flavor:
find . -type f -name \*\(\*\).txt -print0 | xargs -0 sh -c '
for filename ; do
basename_core="${filename##*(}"
basename_core="${basename_core%%)*}"
mv "${filename}" "${basename_core}".txt
done' dummy
This might work for you (GNU sed and shell);
sed -n 's/.*(\(.*\)).*/mv '\''&'\'' '\''\1.txt'\''/p' *.txt
This will print out a list of move commands, after you have validated they are correct, pipe to shell:
sed -n 's/.*(\(.*\)).*/mv '\''&'\'' '\''\1.txt'\''/p' *.txt | sh
find and mv can handle this, with a bash rematch to find your names;
#!/bin/bash
touch lorem\(12312315\).txt
touch ipsum\(578938-12-315-13-416-4\).txt
touch amet\(ran-dom-guid\).txt
pat=".*\((.*)\).txt"
for f in $(find . -type f -name "*.txt" ); do
if [[ $f =~ $pat ]]; then
mv $f ${BASH_REMATCH[1]}.txt
fi
done
ls *.txt
A for loop and Parameter Expansion.
#!/usr/bin/env bash
for f in *\(*\)*.txt; do
temp=${f#*\(}
value=${temp%\)*}
case $value in
*[!0-9-]*) value="${value//-}";;
esac
echo mv -v "$f" "$value.txt"
done
Remove the echo if you're satisfied with the output, so mv can rename/move the files.
Thank you everyone for the responses! I ended up using a mishmash of your suggestions and doing something else entirely, but I'm posting here for posterity...
The files all had one thing in common, the GUID contained in the filename was also always contained in line 2 of the accompanying file, so I yank lane two, strip out the things that are NOT the guid, and rename the file to that string, for any .xml file in the directory where the script is run.
as such:
for i in ./*xml
do
GUID=`cat "$i" | sed -n '2p' | awk '{print $1}' | sed 's/<id>//g' | sed 's/<\/id>//'`
echo File "|" $i "|" is "|" $GUID
done
In the actual script, I do a MV instead of an ECHO and the files are renamed to the guid.
Hopefully this helps someone else in the future, and yes, I know it's wasteful to call sed three times. If I were better with regular expressions, I'm sure I could get that down to one! :)
Thanks!
I have variables where I store regular expressions and than I want to find files in local folder that match that regex
Basically, I have regex for filename and for its extensions which I want to concatenate and then use them in ereg to search for files
So, here is an example of how it looks like:
declare -r CPMOVE_REGEX="^cpmove-[a-z][a-z0-9]{0,16}"
declare -r TAR_REGEX=".tar$"
and then somewhere in the scriptI do the following.
for b in $(ls -1 | egrep "$CPMOVE_REGEX$TAR_REGEX");
do
backups+=($b);
done
let's assume backups variable is declared somewhere before.
I understand that $CPMOVE_REGEX$TAR_REGEX is wrong, and I would like to know how to do it a right way.
One of the ways to make it work is to put your vars in to curly brackets: "${CPMOVE_REGEX}${TAR_REGEX}"
Actually, I found it before posting this message, but I named my test files incorrectly, so they didn't match the regexp. So, that was the issue.
Thanks to Micha Wiedenmann who provided the link to the answer
Your regexp's problem is not in the extension part ($TAR_REGEX) but in the first part. Try to echo the whole string $CPMOVE_REGEX$TAR_REGEX and do some testing on the command line. When the regexp is ok, then put it in the script.
There are no problems combining two variables into one string, and use it as egrep parameter, like in this working example:
root#localhost ~ $ declare -r P1='^[0-9]+\-.{3}'
root#localhost ~ $ ls | egrep "$P1"
20190315-GO2.sql
root#localhost ~ $ declare -r P2='.sql$'
root#localhost ~ $ ls | egrep "$P1$P2"
20190315-GO2.sql
root#localhost ~ $
how to do it a right way.
Probably something along:
declare -r CPMOVE_REGEX="cpmove-[a-z][a-z0-9]{0,16}"
declare -r TAR_REGEX=".tar"
IFS= readarray -d '' -t backups < <(find . -regextype posix-egrep -regex ".*/$CPMOVE_REGEX$TAR_REGEX" -print0)
Pre bashv4 doesn't has -d '' option with readarray. If you bash lower then v4, you need a while read loop:
while IFS= read -r tmp; do
backups+=("$tmp")
done < <(find . -regextype posix-egrep -regex ".*/$CPMOVE_REGEX$TAR_REGEX" -print0)
But because your filenames have no whitespaces in them, which you filter with the regex, then if you are interested in filenames only without the directory paths, probably you could go with reading a newline separated list. That's only if you are sure you will never have any newlines in filenames.
IFS=$'\n' read -r -a backups < <(find . -regextype posix-egrep -regex ".*/$CPMOVE_REGEX$TAR_REGEX" -print "%f\n")
or
IFS=$'\n' backups=($(find . -regextype posix-egrep -regex ".*/$CPMOVE_REGEX$TAR_REGEX" -print "%f\n"))
Notes:
Do not parse ls output. It is not a tool to use in scripts. ls is for pretty printing in your console. Use find. Why you shouldn't parse the output of ls(1). Forget ls in scripts.
Doing for i in $( .. ) is error-prone, the $(...) undergoes word splitting. Although I think you are safe in your script, as you can do grep -x -E "<expression without whitespaces>", it's still conceptually better to use a while read loop. How can I read a file (data stream, variable) line-by-line (and/or field-by-field)?
egrep is deprecated, see the man page. Use grep -E instead.
Quote your variables. Don't backups+=($b) do backups+=("$b"). You will save yourself a lot of troubles later when you just quote all $ expansions. bashfaq quotes
I have 1000 files with following names:
something-345-something.txt
something-5468-something.txt
something-100-something.txt
something-6200-something.txt
and a lot more...
And I have one txt file, with only numbers in it. f.e:
1000
500
5468
6200
699
usw...
Now I would like to move all files, which have a number in their filenames which is in my txt file.
So in my example above the following files should be moved only:
something-5468-something.txt
something-6200-something.txt
Is there an easy way to achieve this?
What about on the fly moving files by doing this:
for i in `cat you-file.txt`; do
find . -iname "*-$i-*" -exec mv '{}' /target/dir \;
; done
For every line in your text file, the find command will try to find only does matching the pattern *-$i-* (something-6200-something.txt) and move it to your target dir.
Naive implementation: for file in $(ls); do grep $(echo -n $file | sed -nr 's/[^-]*-([0-9]+).*/\1/p') my-one-txt.txt && mv $file /tmp/somewhere; done
In English: For every file in output of ls: parse number part of filename with sed and grep for it in your text file. grep returns a non-zero exit code if nothing is found, so mv is in evaluated in that case.
Script file named move (executable):
#!/bin/bash
TARGETDIR="$1"
FILES=`find . -type f` # build list of files
while read n # read numbers from standard input
do # n contains a number => filter list of files by that number:
echo "$FILES" | grep "\-$n-" | while read f
do # move file that passed the filter because its name matches n:
mv "$f" "$TARGETDIR"
done
done
Use it like this:
cd directory-with-files
./move target-directory < number-list.txt
Here's a crazy bit of bash hackery
shopt -s extglob nullglob
mv -t /target/dir *-#($(paste -sd "|" numbers.txt))-*
That uses paste to join all the lines in your numbers file with pipe characters, then uses bash extended pattern matching to find the files matching any one of the numbers.
I assume mv from GNU coreutils for the -t option.
I'm trying to list all PDF files under a given directory $1 (and its subdirectories), get the number of pages in each file and calculate two numbers using the pagecount. My script used to work, but only on filenames that don't contain spaces and only in one directory that is only filled with PDF files. I've modified it a bit already (using quotes around variables and such), but now I'm a bit stuck.
The problem I'm having is that, as it is now, the script only processes the first file found by find . -name '*.pdf'. How would I go about processing the rest?
#!/bin/bash
wd=`pwd`
pppl=0.03 #euro
pppnl=0.033 #eruo
cd $1
for entry in "`find . -name '*.pdf'`"
do
filename="$(basename "$entry")"
pagecount=`pdfinfo "$filename" | grep Pages | sed 's/[^0-9]*//'`
pricel=`echo "$pagecount * $pppl" | bc`
pricenl=`echo "$pagecount * $pppnl" | bc`
echo -e "$filename\t\t$pagecount\t$pricel\t$pricenl"
done
cd "$wd"
The problem with using find in a for loop, is that if you don't quote the command, the filenames with spaces will be split, and if you do quote the command, then the entire results will be parsed in a single iteration.
The workaround is to use a while loop instead, like this:
find . -name '*.pdf' -print0 | while IFS= read -r -d '' entry
do
....
done
Read this article for more discussion: http://mywiki.wooledge.org/ParsingLs
It's a bad idea to use word splitting. Use a while loop instead.
while read -r entry
do
filename=$(basename "$entry")
pagecount=$(pdfinfo "$filename" | grep Pages | sed 's/[^0-9]*//')
pricel=$(echo "$pagecount * $pppl" | bc)
pricenl=$(echo "$pagecount * $pppnl" | bc)
echo -e "$filename\t\t$pagecount\t$pricel\t$pricenl"
done < <(exec find . -name '*.pdf')
Also prefer $() over backticks when possible. You also don't need to place around "" variables or command substitutions when they are being used for assignment.
filename=$(basename "$entry")
As well could simply be just
filename=${entry##*/}
I am trying to create a small Bash script to remove hyphens from a filename. For example, I want to rename:
CropDamageVO-041412.mpg
to
CropDamageVO041412.mpg
I'm new to Bash, so be gentle :] Thank you for any help
Try this:
for file in $(find dirWithDashedFiles -type f -iname '*-*'); do
mv $file ${file//-/}
done
That's assuming that your directories don't have dashes in the name. That would break this.
The ${varname//regex/replacementText} syntax is explained here. Just search for substring replacement.
Also, this would break if your directories or filenames have spaces in them. If you have spaces in your filenames, you should use this:
for file in *-*; do
mv $file "${file//-/}"
done
This has the disadvantage of having to be run in every directory that contains files you want to change, but, like I said, it's a little more robust.
FN=CropDamageVO-041412.mpg
mv $FN `echo $FN | sed -e 's/-//g'`
The backticks (``) tell bash to run the command inside them and use the output of that command in the expression. The sed part applies a regular expression to remove the hyphens from the filename.
Or to do this to all files in the current directory matching a certain pattern:
for i in *VO-*.mpg
do
mv $i `echo $i | sed -e 's/-//g'`
done
A general solution for removing hyphens from any string:
$ echo "remove-all-hyphens" | tr -d '-'
removeallhyphens
$
f=CropDamageVO-041412.mpg
echo "${f//-}"
or, of course,
mv "$f" "${f//-}"