I have thousands of files named "DOCUMENT.PDF" and I want to rename them based on a numeric identifier in the path. Unfortunately, I don't seem to have access to the rename command.
Three examples:
/000/000/002/605/950/ÐÐ-02605950-00001/DOCUMENT.PDF
/000/000/002/591/945/ÐÐ-02591945-00002/DOCUMENT.PDF
/000/000/002/573/780/ÐÐ-02573780-00002/DOCUMENT.PDF
To be renamed as, without changing their parent directory:
2605950.pdf
2591945.pdf
2573780.pdf
Use a for loop, and then use the mv command
for file in *
do
num=$(awk -F "/" '{print $(NF-1)}' file.txt | cut -d "-" -f2);
mv "$file" "$num.pdf"
done
You could do this with globstar in Bash 4.0+:
cd _your_base_dir_
shopt -s globstar
for file in **/DOCUMENT.PDF; do # loop picks only DOCUMENT.PDF files
# here, we assume that the serial number is extracted from the 7th component in the directory path - change it according to your need
# and we don't strip out the leading zero in the serial number
new_name=$(dirname "$file")/$(cut -f7 -d/ <<< "$file" | cut -f2 -d-).pdf
echo "Renaming $file to $new_name"
# mv "$file" "$new_name" # uncomment after verifying
done
See this related post that talks about a similar problem: How to recursively traverse a directory tree and find only files?
Related
I have been trying to get to count all the empty folders in a certain directory. sub-directories excluded. i used the code below but i don't know how to define empty folders or folders that contain files.
echo "$(ls -l | egrep -l $1/* | wc -l)"
the $1 will be the user argument in the command line. example: ./script.sh ~/Desktop/backups/March2021.
Edit - im not allowed to use find command
Edit 2 - ls -l * | awk '/total 0/{print last}{last=$0}' | wc -l this script works but lists all folders even if the directory contains files and data or if the directory is empty.
What about this:
grep -v "." *
I mean the following: "." means any character (I'm not sure the syntax is correct), so basically you look for every file which not even contain any character.
You should not parse ls (directories or file names with newlines), so this solution is only for the assignment:
ls -d */ */* | cut -d/ -f1 | sort | uniq -u | wc -l
Explanation:
ls -d */ shows all directories. This is combined with ls -d */* which will also show contents in the directories.
The resulting output will show all directories.
Empty directories will be shown only once, so you want to look for unique lines.
With the cut you only see the name of the directory, not the files in the directory.
The sort could be skipped here, the ls will give sorted output. When you change the solution to find (next assignment?) the sort might be needed.
uniq can look for lines that occur once. The flag -u removes all lines that have duplicates, so it will show the unique lines in the output.
I'm trying to list some files, but I only want the file names, in order of file date. I've tried a few commands but they don't see to work.
I know that using this code I can list only the file names:
ls -f *
And I know that using this command I can list the files sorted by date:
ls -ltr *
So I have tried using this command to list the file names only, sorted by file date, but it doesn't sort by date:
ls -ltr -f *
That last command simply lists the file names, but sorted by file name, not date.
Any ideas how I can do this with a simple ls command?
FYI, once I get this working my ultimate goal is to only list the most recently created 10 file names, using something like this:
ls -ltr -f * | tail -10
You could try the following command:
ls -ltr | awk '{ print $9 }' | tail -n +2
It extracts the file names from the ls -ltr command.
According to the manual for ls, the -f flag is used to,
-f do not sort, enable -aU, disable -ls --color
One way of extracting only files would be,
ls -p | grep -v /
The option -p is used to append a '/' to a directory name, we can grep for lines not containing a '/'.
To extract 10 most recently used files you could do the following
ls -ptr * | grep -v / | tail -10
My UNIX is quite rusty but what I want is to search a location in UNIX for files containing the two separate words in their text of "generate" and "process", but both words on the SAME LINE?
I know there are script files that contain details of the script author and its function noted at the top of the script. For example, the start of one such script contains the following;
function: generate sales overtime process
I have tried things like the following (again my UNIX is rusty)
grep -rwl . -e "generate" | "process"
But this gives errors such unrecognised commands
What I want is a list of Progress files like;
salesovertime1.p
salestravel1.p
salesexpenses1.p
salesexpenses2.p
If you search for file then find is appropriate, and then you may filter with grep:
find . -exec grep -H generate {} \; 2> /dev/null | grep process
will find recursively every file form the current directory, then filter ones that contains word "generate" and the filter again with ones that contains word "process". Filenames will be produced on output with option -H (GNU grep) and error messages begin redirected to /dev/null.
Now if you want filenames only, you can use :
find . -exec grep -H generate {} \; 2> /dev/null | grep process | cut -f1 -d\:
If you want "generate" and "process" in the same file but on different lines, the following will do it:
grep process `find . -exec grep -H generate {} \; 2> /dev/null | cut -f1 -d\:` 2> /dev/null | cut -f1 -d\:
The find generates a file list that is used to grep against and extract filenames with redirecting errors again.
How can i rename a directory by interchanging the digits and word in directory name.
e.g.
FRA-DEV_007583-K4C-rdf-1
FRA-DEV_007583-K4C-source-8
FRA-DEV_007584-K4C-rdf-19
FRA-DEV_007584-K4C-rdf-8
output should be
FRA-DEV_007583-K4C-1-rdf
FRA-DEV_007583-K4C-8-source
FRA-DEV_007584-K4C-9-rdf
FRA-DEV_007584-K4C-8-rdf
If you have all those files in the same directory, with no other files in there, you could use this script:
#! /bin/bash
nums=(`ls $1 | cut -d- -f5`)
words=(`ls $1 | cut -d- -f4`)
files=(`ls $1 | cut -d- -f1-3`)
complete_files=(`ls $1`)
len=${#complete_files[#]}
for (( i=0; i<${len}; i++ ));
do
newname=${files[$i]}-${nums[$i]}-${words[$i]}
mv $1${complete_files[$i]} $1$newname
done
Save this script as rename.sh in a directory OUTSIDE of the one where your files are. Then execute: bash rename.sh path/to/your/files/ don't forget the final slash, and make a backup first just in case.
how can I remove the word "myfile" in a list of filenames with this structure?
mywork_myfile_XSOP.txt
mywork_myfile_ATTY.txt
mywork_myfile_ATPY.txt
Desired_output:
mywork_XSOP.txt
mywork_ATTY.txt
mywork_ATPY.txt
The simplest method is to use the common rename command which is available in most Unices.
rename 's/^mywork_myfile_/mywork_/' *
This of course expects you to be on the directory of the files. This will not overwrite files. If you want that, just pass the -f option. Also, take note that there's multiple versions of rename out there which may have different options.
Based on this answer on "Rename all files in "Rename all files in directory from $filename_h to $filename_half?", this can be a way:
for file in mywork_myfile*txt
do
mv "$file" "${file/_myfile/}"
done
Note that it uses the bash string operations as follows:
$ file="mywork_myfile_XSOP.txt"
$ echo ${file/_myfile/}
mywork_XSOP.txt
This would work in any Posix shell...
#!/bin/sh
for i
in mywork_myfile_XSOP.txt \
mywork_myfile_ATTY.txt \
mywork_myfile_ATPY.txt; do
set -x
mv "$i" "$(echo $i | sed -e s/myfile_//)"
set +x
done