EX:
$ progaddtext file1 .txt<br>
should rename file1 to file1.txt
I found this code to successfully remove the text from the end of the first arg.
mv $1 $(echo $1 | sed "s/$2$//")
Just having a little trouble figuring out how to do the reverse.
If you have $1 as file1 and $2 as .txt why not just do:
mv "$1" "$1$2"
mv $1 $(echo $1 | sed "s/$/$2/")
Even though using mmv or rename would be much easier, and this question seems to lack a homework tag?
Try:
mv $1 $(echo $1 | sed -e "s/$/.txt/")
Related
I wrote a script which removes spaces in a single folder/file name. I want to make it work so that it removes all spaces in folder/files name in the directory the script exists.
MY Script:
#!/bin/bash
var=$(ls | grep " ")
test=$(echo $var | sed 's/ //')
mv "$var" $test
How it worked
Thank you for helping!
Try this
ls | grep " " | while read file_name
do
mv "$file_name" "$(echo $file_name | sed -E 's/ +//g')"
done
sed -E is so that you can use some simple regex, and / +/ so it can work in case of multiple consecutive spaces such as . And /g so it replaces every occurrences such as foo baa .txt .
Something like this might work:
for f in * ; do
if [[ "$f" =~ \ ]] ; then
mv "$f" "${f// /_}"
fi
done
Explanattion:
for f in * ; do
loops over all file names in the directory. It doesn't have the quirks of ls that make parsing the output of ls a bad idea.
if [[ "$f" =~ \ ]] ; then
This is the bash way of pattern matching. The \ is the pattern. You need to escape the space with a backslash, otherwise the shell will not recognize it as a pattern.
mv "$f" "${f// /_}"
${f// /_} is the bash way of pattern-substitution. The // means replace all occurrences. The syntax is ${variable//pattern/replacement} to replace all patterns in the variable with the replacement.
I have file names (from image tiles) consisting of two numbers separated by an underscore, e.g.
142_27.jpg
7_39.jpg
1_120.jpg
How can I (in linux) add leading zeros to both of these numbers? What I want is the file names as
142_027.jpg
007_039.jpg
001_120.jpg
You can use a single awk command to format filenames with leading zeroes using a printf:
for f in *.jpg; do
echo mv "$f" $(awk -F '[_.]' '{printf "%03d_%03d.%s", $1, $2, $3}' <<< "$f")
done
This will output:
mv 142_27.jpg 142_027.jpg
mv 1_120.jpg 001_120.jpg
mv 7_39.jpg 007_039.jpg
Once you're satisfied with the output, remove echo before mv command.
With perl based rename command
$ touch 142_27.jpg 7_39.jpg 1_120.jpg
$ rename -n 's/\d+/sprintf "%03d", $&/ge' *.jpg
rename(1_120.jpg, 001_120.jpg)
rename(142_27.jpg, 142_027.jpg)
rename(7_39.jpg, 007_039.jpg)
The -n option is for dry run, remove it for actual renaming
If perl based rename command is not available:
$ for f in *.jpg; do echo mv "$f" "$(echo "$f" | perl -pe 's/\d+/sprintf "%03d", $&/ge')"; done
mv 1_120.jpg 001_120.jpg
mv 142_27.jpg 142_027.jpg
mv 7_39.jpg 007_039.jpg
Change echo mv to just mv once dry run seems okay
You can do it with a little shell-script using sed:
for i in *.jpg;
do
new=`echo "$i" | sed -n -e '{ s/^\([0-9]\{0,3\}\)_\([0-9]\{0,3\}\).jpg/000\1_000\2.jpg/ };
{s/\([0-9]*\)\([0-9]\{3\}\)_\([0-9]*\)\([0-9]\{3\}\).jpg/\2_\4.jpg/p };'`;
mv "$i" "$new";
done;
I first append three leading zeros at the said places by default and afterwards cut off as many digits as necessary brginning at the start at said places so that only 3 digits are left
with bash substitution(a,b)
windows(bash)
for f in *.jpg;do a=${f%_*};b=${f#*_};mv $f $(printf "%03d_%07s" $a $b);done
linux
for f in *.jpg;do a=${f%_*};b=${f#*_};b=${b%.*};mv $f $(printf "%03d_%03d".jpg $a $b);done
For ex the file is this:
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
I want to rename this file to:
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN
Using ${parameter%word} (Remove matching suffix pattern):
$ echo "$fn"
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
$ echo "${fn%:*}"
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN
Using cut
$ echo $fn
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
$ echo $fn |cut -d: -f1
NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN
Using awk
echo $fn |awk -F : '{print $1}'
more ways...
According to the link here:
This should work:
awk '{old=$0;gsub(/...$/,"",$0);system("mv \""old"\" "$0)}'
provided the file name is given as input.
For eg:
ls -1 NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00|nawk '{old=$0;gsub(/...$/,"",$0);system("mv \""old"\" "$0)}'
Rename file using bash string manipulations:
# Filename needs to be in a variable
file=NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00
# Rename file
mv "$file" "${file%???}"
This removes the last three characters from filename.
Using just bash:
fn='NBDG6_CDRCCN_4004_-TTNBDG6_CCN_51-140108-1433-802580.00.Blk32768Blk.CCN:00'
mv "$fn" "${fn::-3}"
if you have Ruby
echo NBDG6_CD* | ruby -e 'f=gets.chomp;File.rename(f, f[0..-4])'
Here's what output looks like, basically:
? RESTRequestParamObj.cpp
? plugins/dupfields2/_DupFields.cpp
? plugins/dupfields2/_DupFields.h
I need to get the filenames from second column and pass them to rm. There's AWK script that goes like awk '{print $2}' but I was wondering if there's another solution.
If you have spaces between the ? and the filename then:
cut -c9-
If they're tabs then:
cut -f2
Placed your output in file
$> cat ./text
? RESTRequestParamObj.cpp
? plugins/dupfields2/_DupFields.cpp
? plugins/dupfields2/_DupFields.h
Edit it with sed
$> cat ./text | sed -r -e 's/(\?[\ \t]*)(.*)/\2/g'
RESTRequestParamObj.cpp
plugins/dupfields2/_DupFields.cpp
plugins/dupfields2/_DupFields.h
Sed in here is matching 2 parts of line -
? with tabs or spaces
Other characters until the end f the line
And then it changes whole line only with second part.
This might work for you:
echo "? RESTRequestParamObj.cpp" | sed -e 's/^\S\+/rm /' | sh
or using GNU sed
echo "? RESTRequestParamObj.cpp"| sed -r 's/^\S+/rm /e'
bash only solution, assuming your output comes from stdin:
while read line; do echo ${line##* }; done
use cut/perl instead
cut -f2 -t'\t'|xargs rm -rf
<your output>|perl -ne '#cols = split /\t/; print $cols[1]'|xargs rm -rf
Say I have a directory with a bunch of site names in it.
i.e.
dev.domain.com
dev.domain2.com
dev.domain3.com
How can I rename those to <domain>.com on the linux cli using piping and/or redirection bash?
I get to a point than am stuck.
find . -maxdepth 1 | grep -v "all" | cut --delimiter="." -f3 | awk '{ print $0 }'
Gives me the domain part, but I can't get past that. Not sure awk is the answer either. Any advice is appreciated.
To strip the leading 'dev.' from names it should be like this:
for i in $(find * -maxdepth 1 -type d); do mv $i $(echo $i | sed 's/dev.\(.*\)/\1/'); done
for i in *; do mv $i $( echo $i | sed 's/\([^\.]*\).\([^\.]*\).\([^\.]*\)/\2.\1/' ); done
Explained:
for i in *; do ....; done
do it for every file
echo $i | sed 's/\([^\.]*\).\([^\.]*\).\([^\.]*\)/\2.\1/'
takes three groups of "every character except ." and changes their order
\2.\1 means: print second group, a dot, first group
the $( ... ) takes output of sed and "pastes" it after mv $i and is called "command substitution" http://www.gnu.org/software/bash/manual/bashref.html#Command-Substitution
Try the rename command. It can take a regular expression like this:
rename 's/\.domain.*/.com/' *.com
under the directory you want to work with, try :
ls -dF *|grep "/$"|awk 'BEGIN{FS=OFS="."} {print "mv "$0" "$2,$3}'
will print mv command. if you want to do the rename, add "|sh" and the end:
ls -dF *|grep "/$"|awk 'BEGIN{FS=OFS="."} {print "mv "$0" "$2,$3}'|sh