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//-}"
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 "|" $GUID
done
In the actual script, I do a MV instead of an ECHO and the files are renamed to the guid.
Hopefully this helps someone else in the future, and yes, I know it's wasteful to call sed three times. If I were better with regular expressions, I'm sure I could get that down to one! :)
Thanks!
I am new with Bash, and trying to rename files in my folder keeping the first 9 characters intact and get rid of anything that comes after.
abc123456olda.jpg > abc123456.jpg
I wrote this;
for file in *
do
echo mv "$file" `echo "$file" | sed -e 's/(.{9}).*(\.jpg)$/$1$2/' *.jpg
done
Did not get it to work. Can someone guide what am I doing wrong?
You're not far off, try this:
for file in *.jpg; do
echo mv "$file" "$(echo "$file" | sed -E -e 's/(.{9}).*(\.jpg)$/\1\2/')"
done
There are some corrections. A important one is that $1$2 should be \1\2, and you need the -E flag to sed so that it understands the grouping with parenthesis.
Once you see the command is alright, remove the echo from the second line so mv actually gets executed.
Use bash's built-in parameter expansion operator rather than sed.
Also, you should put *.jpg in the for statement, not the sed argument; what you're doing is processing the contents of the files, not the filenames.
for file in *.jpg
do
mv "$file" "${file:0:9}.jpg"
done
${file:0:9} means the substring of $file starting from index 0 and having 9 characters.
I have a list of filenames like this in bash
UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R1.fq.gz
UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R2.fq.gz
UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz
UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz
UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R1.fq.gz
UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R2.fq.gz
And I want them to look like this
UTSHoS10_R1.fq.gz
UTSHoS10_R2.fq.gz
UTSHoS11_R1.fq.gz
UTSHoS11_R2.fq.gz
UTSHoS12_R1.fq.gz
UTSHoS12_R2.fq.gz
I do not have the perl rename command and sed 's/_Other*160418./_/' *.gz
is not doing anything. I've tried other rename scripts on here but either nothing occurs or my shell starts printing huge amounts of code to the console and freezes.
This post (Removing Middle of Filename) is similar however the answers given do not explain what specific parts of the command are doing so I could not apply it to my problem.
Parameter expansions in bash can perform string substitutions based on glob-like patterns, which allows for a more efficient solution than calling an extra external utility such as sed in each loop iteration:
for f in *.gz; do echo mv "$f" "${f/_Other_*-TTAGGA_R_160418./_}"; done
Remove the echo before mv to perform actual renaming.
You can do something like this in the directory which contains the files to be renamed:
for file_name in *.gz
do
new_file_name=$(sed 's/_[^.]*\./_/g' <<< "$file_name");
mv "$file_name" "$new_file_name";
done
The pattern (_[^.]*\.) starts matching from the FIRST _ till the FIRST . (both inclusive). [^.]* means 0 or more non-dot (or non-period) characters.
Example:
AMD$ ls
UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R1.fq.gz UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R1.fq.gz
UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R2.fq.gz UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R2.fq.gz
UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz
AMD$ for file_name in *.gz
> do new_file_name=$(sed 's/_[^.]*\./_/g' <<< "$file_name")
> mv "$file_name" "$new_file_name"
> done
AMD$ ls
UTSHoS10_R1.fq.gz UTSHoS10_R2.fq.gz UTSHoS11_R2.fq.gz UTSHoS12_R1.fq.gz UTSHoS12_R2.fq.gz
Pure Bash, using substring operation and assuming that all file names have the same length:
for file in UTS*.gz; do
echo mv -i "$file" "${file:0:9}${file:38:8}"
done
Outputs:
mv -i UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R1.fq.gz UTSHoS10_R1.fq.gz
mv -i UTSHoS10_Other_CAAGCC-TTAGGA_R_160418.R2.fq.gz UTSHoS10_R2.fq.gz
mv -i UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz UTSHoS11_R2.fq.gz
mv -i UTSHoS11_Other_AGGCCT-TTAGGA_R_160418.R2.fq.gz UTSHoS11_R2.fq.gz
mv -i UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R1.fq.gz UTSHoS12_R1.fq.gz
mv -i UTSHoS12_Other_GGCAAG-TTAGGA_R_160418.R2.fq.gz UTSHoS12_R2.fq.gz
Once verified, remove echo from the line inside the loop and run again.
Going with your sed command, this can work as a bash one-liner:
for name in UTSH*fq.gz; do newname=$(echo $name | sed 's/_Other.*160418\./_/'); echo mv $name $newname; done
Notes:
I've adjusted your sed command: it had an * without a preceeding . (sed takes a regular expression, not a globbing pattern). Similarly, the dot needs escaping.
To see if it works, without actually renaming the files, I've left the echo command in. Easy to remove just that to make it functional.
It doesn't have to be a one-liner, obviously. But sometimes, that makes editing and browsing your command-line history easier.
I have the following files in the following format:
$ ls CombinedReports_LLL-*'('*.csv
CombinedReports_LLL-20140211144020(Untitled_1).csv
CombinedReports_LLL-20140211144020(Untitled_11).csv
CombinedReports_LLL-20140211144020(Untitled_110).csv
CombinedReports_LLL-20140211144020(Untitled_111).csv
CombinedReports_LLL-20140211144020(Untitled_12).csv
CombinedReports_LLL-20140211144020(Untitled_13).csv
CombinedReports_LLL-20140211144020(Untitled_14).csv
CombinedReports_LLL-20140211144020(Untitled_15).csv
CombinedReports_LLL-20140211144020(Untitled_16).csv
CombinedReports_LLL-20140211144020(Untitled_17).csv
CombinedReports_LLL-20140211144020(Untitled_18).csv
CombinedReports_LLL-20140211144020(Untitled_19).csv
I would like this part removed:
20140211144020 (this is the timestamp the reports were run so this will vary)
and end up with something like:
CombinedReports_LLL-(Untitled_1).csv
CombinedReports_LLL-(Untitled_11).csv
CombinedReports_LLL-(Untitled_110).csv
CombinedReports_LLL-(Untitled_111).csv
CombinedReports_LLL-(Untitled_12).csv
CombinedReports_LLL-(Untitled_13).csv
CombinedReports_LLL-(Untitled_14).csv
CombinedReports_LLL-(Untitled_15).csv
CombinedReports_LLL-(Untitled_16).csv
CombinedReports_LLL-(Untitled_17).csv
CombinedReports_LLL-(Untitled_18).csv
CombinedReports_LLL-(Untitled_19).csv
I was thinking simply along the lines of the mv command, maybe something like this:
$ ls CombinedReports_LLL-*'('*.csv
but maybe a sed command or other would be better
rename is part of the perl package. It renames files according to perl-style regular expressions. To remove the dates from your file names:
rename 's/[0-9]{14}//' CombinedReports_LLL-*.csv
If rename is not available, sed+shell can be used:
for fname in Combined*.csv ; do mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" ; done
The above loops over each of your files. For each file, it performs a mv command: mv "$fname" "$(echo "$fname" | sed -r 's/[0-9]{14}//')" where, in this case, sed is able to use the same regular expression as the rename command above. s/[0-9]{14}// tells sed to look for 14 digits in a row and replace them with an empty string.
Without using an other tools like rename or sed and sticking strictly to bash alone:
for f in CombinedReports_LLL-*.csv
do
newName=${f/LLL-*\(/LLL-(}
mv -i "$f" "$newName"
done
for f in CombinedReports_LLL-* ; do
b=${f:0:20}${f:34:500}
mv "$f" "$b"
done
You can try line by line on shell:
f="CombinedReports_LLL-20140211144020(Untitled_11).csv"
b=${f:0:20}${f:34:500}
echo $b
You can use the rename utility for this. It uses syntax much like sed to change filenames. The following example (from the rename man-page) shows how to remove the trailing '.bak' extension from a list of backup files in the local directory:
rename 's/\.bak$//' *.bak
I'm using the advice given in the top response and have put the following line into a shell script:
ls *.nii | xargs rename 's/[f_]{2}//' f_0*.nii
In terminal, this line works perfectly, but in my script it will not execute and reads * as a literal part of the file name.
I run a script which generated about 10k files in a directory. I just discovered that there is a bug in the script which causes some filenames to have a carriage return (presumably a '\n' character).
I want to run a sed command to remove the carriage return from the filenames.
Anyone knows which params to pass to sed to clean up the filenames in the manner described?
I am running Linux (Ubuntu)
I don't know how sed would do this, but this python script should do the trick:.
This isn't sed, but I find python a lot easier to use when doing things like these:
#!/usr/bin/env python
import os
files = os.listdir('.')
for file in files:
os.rename(file, file.replace('\r', '').replace('\n', ''))
print 'Processed ' + file.replace('\r', '').replace('\n', '')
It strips any occurrences of both \r and \n from all of the filenames in a given directory.
To run it, save it somewhere, cd into your target directory (with the files to be processed), and run python /path/to/the/file.py.
Also, if you plan on doing more batch renaming, consider Métamorphose. It's a really nice and powerful GUI for this stuff. And, it's free!
Good luck!
Actually, try this: cd into the directory, type in python, and then just paste this in:
exec("import os\nfor file in os.listdir('.'):\n os.rename(file, file.replace('\\r', '').replace('\\n', ''))\n print 'Processed ' + file.replace('\\r', '').replace('\\n', '')")
It's a one-line version of the previous script, and you don't have to save it.
Version 2, with space replacement powers:
#!/usr/bin/env python
import os
for file in os.listdir('.'):
os.rename(file, file.replace('\r', '').replace('\n', '').replace(' ', '_')
print 'Processed ' + file.replace('\r', '').replace('\n', '')
And here's the one-liner:
exec("import os\nfor file in os.listdir('.'):\n os.rename(file, file.replace('\\r', '').replace('\\n', '')replace(' ', '_'))\n print 'Processed ' + file.replace('\\r', '').replace('\\n', '');")
If there are no spaces in your filenames, you can do:
for f in *$'\n'; do mv "$f" $f; done
It won't work if the newlines are embedded, but it will work for trailing newlines.
If you must use sed:
for f in *$'\n'; do mv "$f" "$(echo "$f" | sed '/^$/d')"; done
Using the rename Perl script:
rename 's/\n//g' *$'\n'
or the util-linux-ng utility:
rename $'\n' '' *$'\n'
If the character is a return instead of a newline, change the \n or ^$ to \r in any places they appear above.
The reason you aren't getting any pure-sed answers is that fundamentally sed edits file contents, not file names; thus the answers that use sed all do something like echo the filename into a pipe (pseudo file), edit that with sed, then use mv to turn that back into a filename.
Since sed is out, here's a pure-bash version to add to the Perl, Python, etc scripts you have so far:
killpattern=$'[\r\n]' # remove both carriage returns and linefeeds
for f in *; do
if [[ "$f" == *$killpattern* ]]; then
mv "$f" "${f//$killpattern/}"
fi
done
...but since ${var//pattern/replacement} isn't available in plain sh (along with [[...]]), here's a version using sh-only syntax, and tr to do the character replacement:
for f in *; do
new="$(printf %s "$f" | tr -d "\r\n")"
if [ "$f" != "$new" ]; then
mv "$f" "$new"
fi
done
EDIT: If you really want it with sed, take a look at this:
http://www.linuxquestions.org/questions/programming-9/merge-lines-in-a-file-using-sed-191121/
Something along these lines should work similar to the perl below:
for i in *; do echo mv "$i" `echo "$i"|sed ':a;N;s/\n//;ta'`; done
With perl, try something along these lines:
for i in *; do mv "$i" `echo "$i"|perl -pe 's/\n//g'`; done
This will rename all files in the current folder by removing all newline characters from them. If you need to go recursive, you can use find instead - be aware of the escaping in that case, though.
In fact there is a way to use sed:
carr='\n' # specify carriage return
files=( $(ls -f) ) # array of files in current dir
for i in ${files[#]}
do
if [[ -n $(echo "$i" | grep $carr) ]] # filenames with carriage return
then
mv "$i" "$(echo "$i" | sed 's/\\n//g')" # move!
fi
done
This actually works.