How to rename files without changing extension in Linux \
102221.pdf to 102221_name.pdf
This is what you want I think:
for x in *; do mv "$x" "${x%.*}_name.${x##*.}"; done
${x%.*} will give the name of the file without extention
${x##*.} will extract the extentions
ls * | sed -r 'p;s/\.pdf$/_name\.pdf/g' | xargs -n2 mv
list all the files with ls and pipe the output to sed. sed replaces .pdf with _name.pdf and outputs both the original file name and the new file name to xargs with will call mv with the 2 parameters.
you can also use the rename command which is simpler
rename 's/\.pdf$/_name\.pdf/g' ./*
The regex pattern remains the same though
well i am not so good in linux.. but still found a working answer for you.. hope it will solve ur purpose..
check the given link.. you might need a light weighted tool called as jhead mainly its to get the header information about the file link created date and time and other.. you can find the information which suits you..
Answer
https://superuser.com/questions/90057/linux-rename-file-but-keep-extension
jhead
http://www.sentex.net/~mwandel/jhead/
Related
After an HD problem and some work, I have a bunch of files with names like "f1234", "f1235", etc.
My goal is to sort this files according to their filetype. For example, I want to move all the PDF files in the "pdfs" directory.
For one file, I can do : "file f1234", and if it's a PDF, I can "mv f1234 pdfs/". But I have thousands of file... Can you help me with a bash or zsh command for sort all the PDF in one pass ? Thanks
The hard part here is reliably turning the output of file into a directory name. I think probably the best candidate for that is the mime-type of the file rather than the human readable output of file. I'd use something like:
mkdir sorted
for f in f*
do
d=$(file -b --mime-type "$f" | tr / -)
mkdir -p "sorted/$d"
mv "$f" "sorted/$d/"
done
Obviously I'd test that out a bit before running it on your files, but something pretty close to that should work.
I have a list of for example 100 files with the naming convention
<date>_<Time>_XYZ.xml.abc
<date>_<Time>_XYZ.xml
<date>_<Time>_XYZ.csv
for example
20140730_025373_XYZ.xml
20140730_015233_XYZ.xml.ab
20140730_015233_XYZ.csv
Now I want to write script which will remove anything between two underscores. for example in the above case
remove 015233 and change 20140730_015233_XYZ.xml.ab to 20140730_XYZ.xml.ab
remove 015233 and change 20140730_015233_XYZ.csv to 20140730_XYZ.csv
I have tried number of various options using rename, cut, mv but I am getting varied results, not the one which I expect.
You could use rename command if you want to rename files present inside the current directory,
rename 's/^([^_]*)_[^_]*(_.*)$/$1$2/g' *
You can use sed:
sed 's/\([^_]*\)_.*_\(.*\)/\1_\2/' files.list
You can also use cut command
cut -d'_' -f1,3 filename
for FILE in *; do mv "$FILE" "${FILE/_*_/_}"; done
And more specific is
for FILE in *.xml *.xml.ab *.csv; do mv "$FILE" "${FILE/_*_/_}"; done
Further:
for FILE in *_*_*.xml *_*_*.xml.ab *_*_*.csv; do mv "$FILE" "${FILE/_*_/_}"; done
I want to write a script that do specific thing:
I have a txt file e.g.
from1/from2/from3/apple.file;/to1/to2/to3;some not important stuff
from1/from2/banana.file;/to1/to5;some not important stuff
from1/from10/plum.file;/to1//to5/to100;some not important stuff
Now i want to copy file from each line (e.g. apple.file), from original directory tree to new, non existing directories, after first semicolon (;).
I try few code examples from similar questions, but nothing works fine and I'm too weak in bash scripting, to find errors.
Please help :)
need to add some conditions:
file not only need to be copy, but also rename. Example line in file.txt:
from1/from2/from3/apple.file;to1/to2/to3/juice.file;some1
from1/from2/banana.file;to1/to5/fresh.file;something different from above
so apple.file need to be copy and rename to juice.file and put in to1/to2/to3/juice.file
I think thaht cp will also rename file but
mkdir -p "$to"
from answer below will create full folder path with juice.file as folder
In addidtion after second semicolon in each line will be something different, so how to cut it off?
Thanks for all help
EDIT: There will be no spaces in input txt file.
Try this code..
cat file | while IFS=';' read from to some_not_important_stuff
do
to=${to:1} # strip off leading space
mkdir -p "$to" # create parent for 'to' if not existing yet
cp -i "$from" "$to" # option -i to get a warning when it would overwrite something
done
Using awk
(run the awk command first and confirm the output is fine, then add |sh to do the copy)
awk -F";" '{printf "cp %s %s\n",$1,$2}' file |sh
Using shell (get updated that need manually create folder, base on alfe's
while IFS=';' read from to X
do
mkdir -p $to
cp $from $to
done < file
I had this same problem and used tar to solve it! Posted here:
tmpfile=/tmp/myfile.tar
files="/some/folder/file1.txt /some/other/folder/file2.txt"
targetfolder=/home/you/somefolder
tar --file="$tmpfile" "$files"
tar --extract --file="$tmpfile" --directory="$targetfolder"
In this case, tar will automatically create all (sub)folders for you! Best,
Nabi
Sorry if this is very simple compared to usual questions but I am just starting out. I have some files all with the same start name but of different file types, e.g:
1234.x
1234.y
1234.z
1234_V2.x
1234_V2.y
1234_V2.z
I want to rename the first part of these whilst keeping any ending and file type, e.g:
4321.x
4321.y
4321.z
4321_V2.x etc
I have tried using
mv 1234* 4321*
and
rename 1234* 4321*
But no luck! I have also been through all the other SO articles and although I could use a loop, most depend on the file type being the same.
Thanks in advance
You can use bash substitution:
for file in 1234*
do mv "$file" "4321${file#1234}"
done
OR, replace the do mv with the following
do mv "$file" "${file/1234/4321}"
See more in man bash under EXPANSION section, sub-section Parameter Expansion
Assuming your filenames for 1234 and 4321 i.e constant for all files, you can try this
for fn in `find . -name 1234*`
do
newf=`echo $fn | sed s/1234/4321/`
mv $fn $newfn
done
You can use a shell script, but it's kind of ugly because it will fork a lot, and thus, if you have a lot of files to rename, it will take time.
for f in 1234*; do echo mv $f $(echo $f | sed -e 's/1234/4321/'); done
Otherwize, rename is a good way to do it:
rename 's/1234/4321/' 1234*
Rename expects a regular expression as first parameter, see online documentation
See if it works:
rename "s/1234/4321/" 1234*
command means substitute(because of s) occurances of "1234" with "4321" in files that has name of pattern 1234*
You can also look at here. It is slightly more complicated than your case.
I downloaded some files using wget and the files are in the following format:
test.zip?AWSAccesskeyId=XXXXXXX&Expires=00000000&Signature=ZZZZZZZZZZ
Is there any way, to rename those files properly by removing the URL parameters. Also is there any way, to download such files, with proper name(without URL parameter) from wget.
I tried with mmv with the following command:
mmv "*.zip*" "#1.zip"
But I can't find any way to install mmv. I am using CentOS 6. So, please suggest any way, other than this.
for file in *.zip\?*; do mv "$file" "${file%%\?*}"; done
As far as I can tell, there's no option to wget telling it not to include the query string in the local filename. You can use the -O option to specify an explicit filename, and fix the driver script to remove the query string itself.
for i in `ls *.zip?*`; do echo $i | cut -f 1 -d \? | xargs -n1 mv $i ; done
sorry to lazy to do checking for spaces in names right now. and unable to test. i don't have access to a bash based system at this moment in time.
If you have g++ >=4.9.2 then you can install rnm and do:
rnm -ns '/n/.zip' *.zip\?*
# /n/ expands to file name without extension.
Or
rnm -rs '/\.zip\?.*/.zip/' *.zip\?*