I have a bunch of files which are of this format:
blabla.log.YYYY.MM.DD
Where YYYY.MM.DD is something like (2016.01.18)
I have quite a few folders with about 1000 files in each, so I wanted to have a simple script to rename them. I want to rename them to
blabla.log
So basically, I'm just stripping the date at the end. Here is what I have:
for f in [a-zA-Z]*.log.[0-9][0-9][0-9][0-9].[0-9][0-9].[0-9][0-9]; do
mv -v $f ${f#[0-9][0-9][0-9][0-9].[0-9][0-9].[0-9][0-9]};
done
This script outputs this:
mv: `blabla.log.2016.01.18' and `blabla.log.2016.01.18' are the same file
For more information:
I'm on windows, but I run this script in gitbash
For some reason, my gitbash doesn't recognize the "rename" command
Some regex patterns (like [0-9]{4} don't seem to work)
I'm really at a lost. Thanks.
EDIT: I need to rename every single file that has a date at the end and that is of the from: *.log.2016.01.18. They all need to keep their original names. All that should change is the removal of the date.
You have to use % instead of #: you want to remove from the end, not the start of your string.
Also, you're missing a . in what has to be removed, you don't want to end up with blabla.log..
Quoting the variable names prevents surprises when file names contain special characters.
Together:
mv -v "$f" "${f%.[0-9][0-9][0-9][0-9].[0-9][0-9].[0-9][0-9]}"
Related
I've got many files on a linux server which have this format
text_text_mixturelettersnumbers.filefor example Hesperocyparis_goveniana_E00196073A.bam.baior Hesperocyparis_forbesii_RBGEH19_bwa_out.txt. I would like to change the first underscore to a hyphen and leave everything else so it looks like this text-text_mixturelettersnumbers.file.
I have tried rename -n 's/(\w+)_(\w+_.)/$1-$2/' * and many different versions thereof but nothing is happening. Could someone please point out what I've got wrong?
Thanks
Markus
The util-linux rename does not have an option to display the results only. It is very basic.
If you want to list the files that contain two underscores before an extension, use
for f in *_*_*.*; do
echo "$f => ${f/_/-}";
done
To actually rename, use mv:
for f in *_*_*.*; do
mv -- "$f" "${f/_/-}";
done
The "${f/_/-}" replaces the first _ with - in variable f.
Can a bash/shell expert help me in this? Each time I use PDF to split large pdf file (say its name is X.pdf) into separate pages, where each page is one pdf file, it creates files with this pattern
"X 1.pdf"
"X 2.pdf"
"X 3.pdf" etc...
The file name "X" above is the original file name, which can be anything. It then adds one space after the name, then the page number. Page numbers always start from 1 and up to how many pages. There is no option in adobe PDF to change this.
I need to run a shell command to simply remove/strip out all the "X " part, and just leave the digits, like this
1.pdf
2.pdf
3.pdf
....
100.pdf ...etc..
Not being good in pattern matching, not sure what regular expression I need.
I know I need something like
for i in *.pdf; do mv "$i$" ........; done
And it is the ....... part I do not know how to do.
This only needs to run on Linux/Unix system.
Use sed..
for i in *.pdf; do mv "$i" $(sed 's/.*[[:blank:]]//' <<< "$i"); done
And it would be simple through rename
rename 's/.*\s//' *.pdf
You can remove everything up to (including) the last space in the variable with this:
${i##* }
That's "star space" after the double hash, meaning "anything followed by space". ${i#* } would remove up to the first space.
So run this to check:
for i in *.pdf; do echo mv -i -- "$i" "${i##* }" ; done
and remove the echo if it looks good. The -i suggested by Gordon Davisson will prompt you before overwriting, and -- signifies end of options, which prevents things from blowing up if you ever have filenames starting with -.
If you just want to do bulk renaming of files (or directories) and don't mind using external tools, then here's mine: rnm
The command to do what you want would be:
rnm -rs '/.*\s//' *.pdf
.*\s selects the part before (and with) the last white space and replaces it with empty string.
Note:
It doesn't overwrite any existing files (throws warning if it finds an existing file with the target name).
And this operation is failsafe. You can get back the changes made by last rnm command with rnm -u.
Here's a list of documents for rnm.
This should be a basic question for a lot of people, but I am a biologist with no programming background, so please excuse my question.
What I am trying to do is rename about 100,000 gzipped data files that have existing name of a code (example: XG453834.fasta.gz). I'd like to name them to something easily readable and parseable by me (example: Xanthomonas_galactus_str_453.fasta.gz).
I've tried to use sed, rename, and mmv, to no avail. If I use any of those commands on a one-off script then they work fine, it's just when I try to incorporate variables into a shell script do I run into problems. I'm not getting any errors, just no names are changed, so I suspect it's an I/O error.
Here's what my files look like:
#! /bin/bash
# change a bunch of file names
file=names.txt
while IFS=' ' read -r r1 r2;
do
mmv ''$r1'.fasta.gz' ''$r2'.fasta.gz'
# or I tried many versions of: sed -i 's/"$r1"/"$r2"/' *.gz
# and I tried many versions of: rename -i 's/$r1/$r2/' *.gz
done < "$file"
...and here's the first lines of my txt file with single space delimiter:
cat names.txt
#find #replace
code1 name1
code2 name2
code3 name3
I know I can do this with python or perl, but since I'm stuck here working on this particular script I want to find a simple solution to fixing this bash script and figure out what I am doing wrong. Thanks so much for any help possible.
Also, I tried to cat the names file (see comment from Ashoka Lella below) and then use awk to move/rename. Some of the files have variable names (but will always start with the code), so I am looking for a find & replace option to just replace the "code" with the "name" and preserve the file name structure.
I suspect I am not escaping the variable within the single tick of the perl expression, but I have poured over a lot of manuals and I can't find the way to do this.
If you're absolutely sure than the filenames doesn't contain spaces of tabs, you can try the next
xargs -n2 < names.txt echo mv
This is for DRY run (will only print what will do) - if you satisfied with the result, remove the echo ...
If you want check the existence ot the target, use
xargs -n2 < names.txt echo mv -i
if you want NEVER allow overwriting of the target use
xargs -n2 < names.txt echo mv -n
again, remove the echo if youre satisfied.
I don't think that you need to be using mmv, a simple mv will do. Also, there's no need to specify the IFS, the default will work for you:
while read -r src dest; do mv "$src" "$dest"; done < names.txt
I have double quoted the variable names as it is generally considered good practice but in this case, a space in either of the filenames will result in read not working as you expect.
You can put an echo before the mv inside the loop to ensure that the correct command will be executed.
Note that in your file names.txt, the .fasta.gz suffix is already included, so you shouldn't be adding it inside the loop aswell. Perhaps that was your problem?
This should rename all files in column1 to column2 of names.txt. Provided they are in the same folder as names.txt
cat names.txt| awk '{print "mv "$1" "$2}'|sh
I am trying to write a bash shell script to rename a bunch of photos to my own numbering system. All images filenames are like "IMG_0000.JPG" and I can get the script to match and rename(overwrite) all the photos with the following Perl-regex code:
#!/bin/bash
rename -f 's/\w{4}\d{4}.JPG/replacement.jpg/' *.JPG;
But when I try to use a variable as the name of the replacement, as I keep seeing on other posts here and elsewhere on the internet, nothing happens:
#!/bin/bash
$replacement = "000.jpg";
rename -f 's/\w{4}\d{4}.JPG/$replacement/' *.JPG;
How can I get such a variable to work correctly in my bash script? (NOTE: I am not looking to simply strip the "IMG_" from the filename)
Take the replacement out of single quotes:
#!/bin/bash
$replacement="000.jpg"
rename -f 's/\w{4}\d{4}.JPG/'$replacement'/' *.JPG
Bash does not inspect single quoted strings for interpolation.
Using double quotes and correct variable assignment:
#!/bin/bash
replacement="000.jpg"
rename -f "s/\w{4}\d{4}\.JPG/$replacement/" *.JPG
Note that this can cause trouble, e.g. when renaming two files with names like IMG_0001.JPG and FOO_9352.JPG: The first file will be renamed to 000.jpg, then the second file will also be renamed to 000.jpg, overwriting the first.
I've been trying to figure out a command that will search through 13+ files and replace
all matches and variances of forms data and replace them with form data enhancements.
The trick is that there could be [whitespace] - or _ as a separator that I would like
to preserve. I'm running form command line so I believe I could run this script multiple
times and just point it at the file, or if there's a way to capture all files in a directory
(even including directory names) it might just be easier.
I believe its something to the tune of
sed "s/forms_data/form-data-enhancements/g ; s/forms-data/form-data-enhancements/g ; s/forms data/form data enhancements/g" oldfile > newfile
nut I'm not sure.....
variances might be
forms-data
forms_data
forms data
etcetra. Would someone mind sharing a bit of sed awk wisdom? The best I can find is something called an arrary replace but was unable to get any information on how to use it.
Thanks greatly.
Will this work for you -
sed -i 's/\<forms[ -_]data\>/form data enhancements/g' /path/to/files*
-i will do in-file substitution. So first pick a file and run the command without the -i option. If everything looks ok then you can go ahead and use the -i.
Update:
If you would like to retain the separators then you can do something like this -
sed -i 's/\<forms\([ -_]\)data\>/form\1data\1enhancements/' /path/to/files*