I use the following command to rename all files with no spaces adding prefix "Hello"
for FILENAME in *; do mv $FILENAME Hello_$FILENAME; done
I use the following command to get unix datestamp
date +%s
How do I replace Hello with date +%s output?
If I understand your question, then you could use the $(date +%s) command substitution syntax to get the command output (and I suggest quotes) like
for i in *; do mv "$i" "$(date +%s)_$i"; done
Related
I have the output file day by day:
linux-202105200900-foo.direct.tar.gz
The date and time string, ex: 202105200900 will change every day.
I need to manually rename these files to
linux-202105200900x86-foo.direct.tar.gz
( insert a short string x86 after date/time )
any bash script can help to do this?
If you're always inserting the string "x86" at character #18 in the string, you may use that command:
var="linux-202105200900-foo.direct.tar.gz"
var2=${var:0:18}"x86"${var:18}
echo $var2
The 2nd line means: "assign to variable var2 the first 18 characters of var, followed by x86 followed by the rest of the variable var"
If you want to insert "x86" just before the last hyphen in the string, you may write it like this:
var="linux-202105200900-foo.direct.tar.gz"
var2=${var%-*}"x86-"${var##*-}
echo $var2
The 2nd line means: "assign to variable var2:
the content of the variable var after removing the shortest matching pattern "-*" at the end
the string "x86-"
the content of the variable var after removing the longest matching pattern "*-" at the beginning
In addition to the very good answer by #Jean-Loup Sabatier another, perhaps more general way would simply be to replace the second occurrence of '-' with x86- which you can do with sed. Let's say you have:
fname=linux-202105200900-foo.direct.tar.gz
You can update that with:
fname="$(sed 's/-/x86-/2' <<< "$fname")"
Which simply uses a command substitution with sed and a herestring to modify fname assigning the modified result back to fname.
Example Use/Output
$ fname=linux-202105200900-foo.direct.tar.gz
fname="$(sed 's/-/x86-/2' <<< "$fname")"
echo $fname
linux-202105200900x86-foo.direct.tar.gz
Do you need this?
❯ dat=$(date '+%Y%m%d%H%M%S'); echo ${dat}
20210520170336
❯ filename="linux-${dat}x86-foo.direct.tar.gz"; echo ${filename}
linux-20210520170336x86-foo.direct.tar.gz
I wanted to go as simple as possible, considering only the timestamp is going to change, this script should do it. Just run it inside the folder where files are located and you'll get all of them renamed with x86.
#!/bin/bash
for file in $(ls); do
replaced=$(echo $file | sed 's|-foo|x86-foo|g')
mv $file $replaced
done
This is my output
filip#filip-ThinkPad-T14-Gen-1:~/test$ ls
linux-202105200900-foo.direct.tar.gz linux-202105201000-foo.direct.tar.gz linux-202105201100-foo.direct.tar.gz
filip#filip-ThinkPad-T14-Gen-1:~/test$ ./../development/bash-utils/bulk-rename.sh
filip#filip-ThinkPad-T14-Gen-1:~/test$ ls
linux-202105200900x86-foo.direct.tar.gz linux-202105201000x86-foo.direct.tar.gz linux-202105201100x86-foo.direct.tar.gz
Simply iterate through all the files in current folder and pipeline result to sed to replace regex -foo with x86-foo, then rename file with mv command.
As David mentioned in comment, if you're worried that there could be multiple occurrences of -foo then you can just replace g as global to 1 as first occurrence and that's it!
There is also the rename utility (https://man7.org/linux/man-pages/man1/rename.1.html), you could use:
rename -v 0-foo.direct.tar.gz 0x86-foo.direct.tar.gz *
which results in
`linux-202105200900-foo.direct.tar.gz' -> `linux-202105200900x86-foo.direct.tar.gz'
`linux-202205200900-foo.direct.tar.gz' -> `linux-202205200900x86-foo.direct.tar.gz'
`linux-202305200900-foo.direct.tar.gz' -> `linux-202305200900x86-foo.direct.tar.gz'
In addition to the very good answer by #David C. Rankin, just adding it in a loop and renaming the files
# !/usr/bin/bash
for file in `ls linux* 2>/dev/null` # Extract all files starting with linux
do
echo $file
fname="$(sed 's/-/x86-/2' <<< "$file")"
mv "$file" "$fname" # Rename file
done
Output recieved :
linux-202105200900x86-foo.direct.tar.gz
I want to change a filename "Domain_20181012230112.csv" to "Domain_12345_20181012230112.csv" where "Domain" and "12345" are constants while 20181012230112 is always gonna change but with fix length. In bash how can I do this
If all you want is to replace Domain_ with Domain_12345_, then just do
for file in Domain_*;
do
mv "$file" "${file/Domain_/Domain_12345_}"
done
You can make it even shorter if you know that there will only be one underscore:
...
mv "$file" "${file/_/_12345_}"
...
See string substitutions for more info.
You can use mv in a for loop, like this:
for file in Domain_??????????????.csv ; do ts=`echo ${file} | cut -c8-21`; mv ${file} Domain_12345_${ts}.csv; done
Given the one file of your example, this will essentially execute this command
mv Domain_20181012230112.csv Domain_12345_20181012230112.csv
You can simply use the date command to get the date and time information you want
date '+%Y-%m-%d %H:%M:%S'
# 2018-10-26 10:25:47
To then use the result within the filename, you can put it in `` to evaluate it inline, for example you can run
echo "Domain_12345_`date '+%Y-%m-%d %H:%M:%S'`"
# Domain_12345_2018-10-26 10:29:17
You can use the date's man page to figure out the option for milliseconds to add es well.
man date
There are different options like %m and %d for example that always have leading zeroes if necessary, so the file name length stays constant.
To then rename the file you can use the mv (move) command
mv "Domain_20181012230112.csv" "Domain_12345_`date '+%Y-%m-%d %H:%M:%S'`.csv"
Good luck with the rest of the exercise!
i have a directory with a lot of sub-directories including files.
For each WAV file i would like to rename WAV file by adding creation date (date when file WAV has been firstly created) at the beginning of the file (without changing timestamps of file itself).
Next step would be to convert the WAV file to MP3 file, so i will save hard drive space.
for that purpose, i'm trying to create a bash script but i'm having some issues.
I want to keep the same structure as original directory and therefore i was thinking of something like:
for file in `ls -1 *.wav`
do name=`stat -c %y $file | awk -F"." '{ print $1 }' | sed -e "s/\-//g" -e "s/\://g" -e "s/[ ]/_/g"`.wav
cp -r --preserve=timestampcp $dir_original/$file $dir_converted/$name
done
Don't use ls to generate a list of file names, just let the shell glob them (that's what ls *.wav does anyway):
for file in ./*.wav ; do
I think you want the timestamp in the format YYYYMMDD_HHMMSS ?
You could use GNU date with stat to have a somewhat neater control of the output format:
epochtime=$(stat -c %Y "$file" )
name=$(date -d "#$epochtime" +%Y%m%d_%H%M%S).wav
stat -c %Y (or %y) gives the last modification date, but you can't really get the date of the file creation on Linux systems.
That cp looks ok, except for the stray cp at the end of timestampcp, but that must be a typo. If you do *.wav, the file names will be relative to current directory anyway, so no need to prefix with $dir_original/.
If you want to walk through a whole subdirectory, use Bash's globstar feature, or find. Something like this:
shopt -s globstar
cd "$sourcedir"
for file in ./**/*.wav ; do
epochtime=$(stat -c %Y "$file" )
name=$(date -d "#$epochtime" +%Y%m%d_%H%M%S).wav
dir=$(dirname "$file")
mkdir -p "$target/$dir"
cp -r --preserve=timestamp "$file" "$target/$dir/$name"
done
The slight inconvenience here is that cp can't create the directories in the path, so we need to use mkdir there. Also, I'm not sure if you wanted to keep the original filename as part of the resulting one, this would remove it and just replace the file names with the timestamp.
I did some experimenting with the calculation of name to see if I could get it more succinctly, and came up with this:
name=$(date "+%Y%m%d_%H%M%S" -r "$file")
I wanted to append all file names in that folder with the date they were created , and below works perfectly.
#############################
#!/bin/sh
for file in `ls *.JPG`;
do
mv -f "$file" "$(date -r "$file" +"%Y%m%d_%H_%M_%S")_"$file".jpg"
done
##############################
I have a folder with lots of files having a pattern, which is some string followed by a date and time:
BOS_CRM_SUS_20130101_10-00-10.csv (3 strings before date)
SEL_DMD_20141224_10-00-11.csv (2 strings before date)
SEL_DMD_SOUS_20141224_10-00-10.csv (3 strings before date)
I want to loop through the folder and extract only the part before the date and output into a file.
Output
BOS_CRM_SUS_
SEL_DMD_
SEL_DMD_SOUS_
This is my script but it is not working
#!/bin/bash
# script variables
FOLDER=/app/list/l088app5304d1/socles/Data/LEMREC/infa_shared/Shell/Check_Header_T24/
LOG_FILE=/app/list/l088app5304d1/socles/Data/LEMREC/infa_shared/Shell/Check_Header_T24/log
echo "Starting the programme at: $(date)" >> $LOG_FILE
# Getting part of the file name from FOLDER
for file in `ls $FOLDER/*.csv`
do
mv "${file}" "${file/date +%Y%m%d HH:MM:SS}" 2>&1 | tee -a $LOG_FILE
done #> $LOG_FILE
Use sed with extended-regex and groups to achieve this.
cat filelist | sed -r 's/(.*)[0-9]{8}_[0-9][0-9]-[0-9][0-9].[0-9][0-9].csv/\1/'
where filelist is a file with all the names you care about. Of course, this is just a placeholder because I don't know how you are going to list all eligible files. If a glob will do, for example, you can do
ls mydir/*.csv | sed -r 's/(.*)[0-9]{8}_[0-9][0-9]-[0-9][0-9].[0-9][0-9].csv/\1/'
Assuming you wont have numbers in the first part, you could use:
$ for i in *csv;do str=$(echo $i|sed -r 's/[0-9]+.*//'); echo $str; done
BOS_CRM_SUS_
SEL_DMD_
SEL_DMD_SOUS_
Or with parameter substitution:
$ for i in *csv;do echo ${i%_*_*}_; done
BOS_CRM_SUS_
SEL_DMD_
SEL_DMD_SOUS_
When you use ${var/pattern/replace}, the pattern must be a filename glob, not command to execute.
Instead of using the substitution operator, use the pattern removal operator
mv "${file}" "${file%_*-*-*.csv}.csv"
% finds the shortest match of the pattern at the end of the variable, so this pattern will just match the date and time part of the filename.
The substitution:
"${file/date +%Y%m%d HH:MM:SS}"
is unlikely to do anything, because it doesn't execute date +%Y%m%d HH:MM:SS. It just treats it as a pattern to search for, and it's not going to be found.
If you did execute the command, though, you would get the current date and time, which is also (apparently) not what you find in the filename.
If that pattern is precise, then you can do the following:
echo "${file%????????_??-??-??.csv}" >> "$LOG_FILE"
using grep:
ls *.csv | grep -Po "\K^([A-Za-z]+_)+"
output:
BOS_CRM_SUS_
SEL_DMD_
SEL_DMD_SOUS_
How can I create a subdirectory using BASH shell that is a datetime stamp of the format YYYMMDDHHMMSS? I am using mkdir ~/$(printf "%s" `date +"%Y%m%d%H%M%S"`) but keep getting unprintable characters ('?') on the end. Linux 2.6.18
You're don't need the printf to strip off newlines, $() will do that for you (see bash manual)
mkdir ~/$(date +"%Y%m%d%H%M%S")
To see what those unprintable chars are (you may have a trailing carriage return), try this:
for dir in 2013*; do printf "%s" "$dir" | od -c; done
What about using :
DIR_NAME="/Your_Path/"mydir`date +"%Y%m%d%H%M%S"`
or
printf `date +"%Y%m%d%H%M%S"`// Displaying 20130429173838