Shell: replace \n with space between two specific lines - shell

This is my first post here, thanks in advance for your prompt support.
I did some greps on a big file and came with file like the below. I want to make all lines after that starting with XX# in the same line.
So, I want the flowing:
XX#DEV1>
Feb 23 07:00:03
Feb 23 07:00:05
Sent : 4608
Received : 4227
Feb 23 07:00:07
Feb 23 07:00:09
XX#DEV2>
Feb 23 07:00:32
Feb 23 07:00:34
Sent : 4608
Received : 4232
Feb 23 07:00:36
Feb 23 07:00:38
XX#DEV1>
Feb 23 08:00:03
Feb 23 08:00:06
Sent : 4608
Received : 4265
Feb 23 08:00:07
Feb 23 08:00:09
XX#DEV2>
...
To become:
XX#DEV1> Feb 23 07:00:03 Feb 23 07:00:05 Sent : 4608 Received : 4227 Feb 23 07:00:07 Feb 23 07:00:09
XX#DEV2> Feb 23 07:00:32 Feb 23 07:00:34 Sent : 4608 Received : 4232 Feb 23 07:00:36 Feb 23 07:00:38
XX#DEV1> Feb 23 08:00:03 Feb 23 08:00:06 Sent : 4608 Received : 4265 Feb 23 08:00:07 Feb 23 08:00:09
XX#DEV2> ...

You can do that with awk pretty easily:
awk '/^XX/{if(length(out))print out;out=$0;next}{out=out" "$0}END{print out}' yourfile
That says.. if the line starts with "XX", print whatever we have accumluated in variable out, then save current line in variable out. If the line is anything else, append it to variable out after adding a space. At the end, print whatever we have accumulated in variable out.
Output:
XX#DEV1> Feb 23 07:00:03 Feb 23 07:00:05 Sent : 4608 Received : 4227 Feb 23 07:00:07 Feb 23 07:00:09
XX#DEV2> Feb 23 07:00:32 Feb 23 07:00:34 Sent : 4608 Received : 4232 Feb 23 07:00:36 Feb 23 07:00:38
XX#DEV1> Feb 23 08:00:03 Feb 23 08:00:06 Sent : 4608 Received : 4265 Feb 23 08:00:07 Feb 23 08:00:09

Or you can do it bash in much the same way as the awk in my other answer:
#!/bin/bash
while read line
do
if [[ $line == XX* ]] ;then
echo $OUT # Output whatever we have accumulated in $OUT
OUT=$line # Zero out $OUT and restart accumulating
else
OUT="$OUT $line" # Append this line to $OUT
fi
done < yourfile
echo $OUT # Output any trailing stuff at the end

sed -n '
# use -n to prevent sed from printing each line
# if the line read starts with XX, go (b) to :printit
/^XX/ b printit
# if it is the last line ($), also go to :printit
$ b printit
# for all other lines, append the read line to the hold space (H)
# and then go to the :end (so we read the next line)
H; b end
:printit {
# swap the hold and the pattern space
x
# replace all (s/../../g) single or multiple newlines (\n\n*)
# with a space
s/\n\n*/ /g
# print the pattern space
p
}
:end
' your_file

Related

bash console output print layout

I want to make my writing to the console output in a nice tidy human readable.
here is how it looks now:
====================== Sat Apr 16 12:57:17 EDT 2022 ======================
==========================================================================
====================== Leopard - Download from S3 ======================
==========================================================================
==========================================================================
====================== Leopard - Decompressing ======================
==========================================================================
total 1349872
drwxr-xr-x 2 root root 12288 Apr 16 12:57 .
drwxrwxrwt. 4 root root 102 Apr 16 12:57 ..
-rw-r--r-- 1 root root 185070885 Apr 16 12:03 asdasdasd.sql.gz
-rw-r--r-- 1 root root 40344632 Apr 16 12:03 asdasdas.sql.gz
-rw-r--r-- 1 root root 26631 Apr 16 12:03 asdad.sql.gz
-rw-r--r-- 1 root root 1679 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1237 Apr 16 12:03 asd.sql.gz
-rw-r--r-- 1 root root 5241900 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1144 Apr 16 12:03 asdasasd.sql.gz
-rw-r--r-- 1 root root 489312 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1138 Apr 16 12:03 asdasdasd.sql.gz
==========================================================================
====================== NewYorkCity - Download from S3 ======================
==========================================================================
==========================================================================
====================== NewYorkCity - Unloading SSL Example ======================
==========================================================================
total 1349872
drwxr-xr-x 2 root root 12288 Apr 16 12:57 .
drwxrwxrwt. 4 root root 102 Apr 16 12:57 ..
-rw-r--r-- 1 root root 185070885 Apr 16 12:03 asdasdasd.sql.gz
-rw-r--r-- 1 root root 40344632 Apr 16 12:03 asdasdas.sql.gz
-rw-r--r-- 1 root root 26631 Apr 16 12:03 asdad.sql.gz
-rw-r--r-- 1 root root 1679 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1237 Apr 16 12:03 asd.sql.gz
-rw-r--r-- 1 root root 5241900 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1144 Apr 16 12:03 asdasasd.sql.gz
-rw-r--r-- 1 root root 489312 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1138 Apr 16 12:03 asdasdasd.sql.gz
I want that all the === line will be in the same length, and the text always in the center with 1 space on each side
Will appreciate assistance here :)
UPDATE / EDIT:
The original script is something like that:
eecho () { echo ==========================================================================; }
echo_stage () {
START=1
END=11
for (( c=$START; c<=$END; c++ ))
do
printf == '-%.0s'
done
echo -n " " $1 " "
for (( c=$START; c<=$END; c++ ))
do
printf == '-%.0s'
done
echo
}
stage() {
eecho
echo_stage "$1" "$2"
eecho
}
print_date () { echo "======================" $(date) "======================"; }
reload_db() {
print_date
rm -rf /var/tmp/db
mkdir -p /var/tmp/db
stage "DB - Download from S3"
aws s3 sync s3://db-backup/latest/ /var/tmp/db --profile=papilon --quiet
stage "DB - Decompressing"
pigz -d /var/tmp/db/*
stage "DB - Restoring Data"
cd /var/tmp/db
stage "DB - Restoring Tables"
for i in `ls -1 *.sql | grep -v "_view.sql"`;do echo $i;mysql db < $i;done
stage "DB - Restoring Views"
for i in `ls -1 *.sql | grep "_view.sql"`;do echo $i;mysql db < $i;done
stage "DB - Clean up"
rm -rf /var/tmp/db
print_date
}
reload_db
This awk filter will size your === padding correctly, and align titles to the center, if you pipe your command output through it:
# cmd |
awk '
BEGIN {a[1] = "="}
NF>1 && $1~/^=+$/ && $NF~/^=+$/ {
sub(/^=+/, "")
sub(/=+$/, "")
title_len=length($0)
pad = ""
for (i=1; i<=(74-title_len)/2; ++i) {
pad=pad"="
}
$0 = pad $0 pad a[title_len%2]
}
1'
This doesn't wrap the ls -l output, only the titles. It assumes 74 is hardcoded as the length of the solid === lines (and doesn't change depending on terminal size). The array a is used to add an extra = when 74 - title_len is an odd number.
Example output:
====================== Sat Apr 16 12:57:17 EDT 2022 ======================
==========================================================================
====================== Leopard - Download from S3 ======================
==========================================================================
==========================================================================
======================= Leopard - Decompressing ========================
==========================================================================
total 1349872
drwxr-xr-x 2 root root 12288 Apr 16 12:57 .
drwxrwxrwt. 4 root root 102 Apr 16 12:57 ..
-rw-r--r-- 1 root root 185070885 Apr 16 12:03 asdasdasd.sql.gz
-rw-r--r-- 1 root root 40344632 Apr 16 12:03 asdasdas.sql.gz
-rw-r--r-- 1 root root 26631 Apr 16 12:03 asdad.sql.gz
-rw-r--r-- 1 root root 1679 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1237 Apr 16 12:03 asd.sql.gz
-rw-r--r-- 1 root root 5241900 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1144 Apr 16 12:03 asdasasd.sql.gz
-rw-r--r-- 1 root root 489312 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1138 Apr 16 12:03 asdasdasd.sql.gz
==========================================================================
==================== NewYorkCity - Download from S3 ====================
==========================================================================
==========================================================================
================= NewYorkCity - Unloading SSL Example ==================
==========================================================================
total 1349872
drwxr-xr-x 2 root root 12288 Apr 16 12:57 .
drwxrwxrwt. 4 root root 102 Apr 16 12:57 ..
-rw-r--r-- 1 root root 185070885 Apr 16 12:03 asdasdasd.sql.gz
-rw-r--r-- 1 root root 40344632 Apr 16 12:03 asdasdas.sql.gz
-rw-r--r-- 1 root root 26631 Apr 16 12:03 asdad.sql.gz
-rw-r--r-- 1 root root 1679 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1237 Apr 16 12:03 asd.sql.gz
-rw-r--r-- 1 root root 5241900 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1144 Apr 16 12:03 asdasasd.sql.gz
-rw-r--r-- 1 root root 489312 Apr 16 12:03 asdasd.sql.gz
-rw-r--r-- 1 root root 1138 Apr 16 12:03 asdasdasd.sql.gz
Because you have now posted your script, I will add a new answer for pure bash.
title() {
local text pad
(( ${#1} > 70 )) && { echo "$1"; return; }
text=${1:+ }$1${1:+ }
pad=$( eval "printf %.1s ={1..$(( ( 74 - ${#text} ) / 2 ))}" )
echo "$pad$text$pad$( (( ${#text} % 2 )) && printf = )"
}
For title 'foo bar', this function prints foo bar (or any string up to 70 characters) with a space either side, centered and padded to 74 columns with =. If the string is longer than 70 characters it's too long to pad, so it's printed as is.
================================ foo bar =================================
With no argument, or an empty argument, it prints a solid line of 74 =:
==========================================================================
You can swap = for any single ASCII character.
You can swap 74 for any even number. Also for an odd number, if you change && printf to || printf in the last line. (also change the 70 to N - 4)
You can call title once for the date, or three times for the larger three line banners (see banner below).
Explanation:
${#text} is bash for "length of $text".
${1:+ } expands to a space, unless $1 is empty or unset. This allows us to add spaces to either end of the string, or exclude them for an empty string.
={1..10} expands to =1 =2 =3 ... =10, and printf %.1s prints the first character of each string. Combining these allows us to repeat a string (=) N times.
But we can't use a variable (or arithmetic) in {1..10} normally. So we need eval.
You probably heard eval is bad, and a security risk. That's often true, but here we are not passing any unknown data to eval (such as user input), and it's safe from code injection. (${#text} always expands to a single number)
So we make two bars, of length (74 - text-length) / 2, adding another = to the second bar if the text length is an odd number.
I also made a few changes to your script which you might consider. Apart from the date and titles, these have nothing to do with the padding. The title and banner functions will work on your old script.
title() {
local text pad
(( ${#1} > 70 )) && { echo "$1"; return; }
text="${1:+ }$1${1:+ }"
pad=$( eval "printf %.1s ={1..$(( ( 74 - ${#text} ) / 2 ))}" )
echo "$pad$text$pad$( (( ${#text} % 2 )) && printf = )"
}
banner() {
title
title "$1"
title
}
reload_db() {
local i
title "$(date)"
rm -rf /var/tmp/db || exit 1
mkdir -p /var/tmp/db || exit 1
banner 'DB - Download from S3'
aws s3 sync s3://db-backup/latest/ /var/tmp/db --profile=papilon --quiet
banner 'DB - Decompressing'
pigz -d /var/tmp/db/*
banner 'DB - Restoring Data'
cd /var/tmp/db || exit 1
banner 'DB - Restoring Tables'
GLOBIGNORE='*_view.sql'
for i in *.sql; do
echo "$i"
mysql db < "$i"
done
GLOBIGNORE=
banner 'DB - Restoring Views'
for i in *_view.sql; do
echo "$i"
mysql db < "$i"
done
banner 'DB - Clean up'
rm -rf /var/tmp/db
title "$(date)"
}
At the very least, you should exit early if cd, mkdir, or the first rm fail. Also, looping over an unquoted ls command sub is a bad idea. Instead you can use glob expansion (or find).
You could also concatenate the SQL scripts, but this only works if all the commands end with a semicolon (see Run multiple sql files in mysql batch):
big_title 'DB - Restoring Tables'
GLOBIGNORE='*_view.sql'
printf '%s\n' *.sql
cat *.sql | mysql db
GLOBIGNORE=
big_title 'DB - Restoring Views'
printf '%s\n' *_view.sql
cat *_view.sql | mysql db

Continuously read the last line of log file in bash script

I have a log file in which new lines are continuously written.
I would like a bash script that continuously reads the last line of this log file, so that I can process the line (e.g. execute a specific command if the line contains the word "error").
I've tried:
while true
do
if tail -n1 -f file.log | grep -q ERROR
then
echo "$(date) : ERROR detected"
fi
done
But it's spamming:
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
sun 21 mar 2021 18:32:41 CET : ERROR detected
(a new line is added every minute in this example)
How can I read only the last line and do not have spam for the result ?
I suggest with GNU grep:
tail -n1 -f file.log | grep --line-buffered ERROR | while read; do echo "$(date) : ERROR detected"; done
This is exactly why tail -f has been invented:
tail -f <logfile>
will show the last line of your logfile, so you can follow what gets added.
This can be combined with a grep:
tail -f <logfile> | grep <text_to_be_searched>
In your case:
tail -f file.log | grep "ERROR"

List all the mondays of this month

I'm pretty new to bash and all the terminal in general - I've been messing around with cal and the date scripts wondering if there is anyway to list all the dates of monday of the current month .
My thought process is going thru the cal command, listing out the dates and maybe cutting a column from that input. Is that possible ?
You can do it with date command. Print 10 mondays since month ago:
for x in $(seq 0 9)
do
date -d "$x monday 5 week ago"
done
And grep only current month. Full command: for x in $(seq 0 9); do; date -d "$x monday 5 week ago"; done | grep $(date +%b)
Output:
Mon Jun 5 00:00:00 MSK 2017
Mon Jun 12 00:00:00 MSK 2017
Mon Jun 19 00:00:00 MSK 2017
Mon Jun 26 00:00:00 MSK 2017
Given:
$ cal
June 2017
Su Mo Tu We Th Fr Sa
1 2 3
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30
You can do:
$ cal | awk 'NF>5{print $2}'
Mo
5
12
19
26
If you want something that will support any day of cal, use the field width (gawk only this):
$ cal | gawk -v n=5 '
BEGIN{
FIELDWIDTHS = "3 3 3 3 3 3 3"
}
FNR>1{print $n}'
Th
1
8
15
22
29
Or, as pointed out in comments:
$ ncal | awk '/^Mo/'
Mo 5 12 19 26
Combination of cal,cut commands to achieve the output.
cal -h| cut -c'4,5'
Remove the highlight and cut the characters which suits in the fields of monday.
ncal | sed -n '/^Mo/p'
The output as below:
Mo 5 12 19 26

Please help me how to delete the files in one folder which are more than 60 days old in UNIX

I know how to delete the files the files which are more than 60 days old. But I have to satisfy below conditions. Please help me to get correct script to automate this.
I have below files for each day on monthly basis. So I have these files for last 3 years.
vtm_data_12month_20140301.txt
vtm_data_12month_20140301.control
vtm_mtd_20130622.txt
vtm_mtd_20130622.control
vtm_ytd_20131031.txtvtm_ytd_20131031.control
I'd like to write a script find the all files which are more than 60 days old and delete them all but except last month file.
Suppose for january I want to keep the last file (latest) vtm_data_12month_20140131.txt and delete all 30 files. Issue here is, there is chance that I might have files received for January 30th, so in that case I should not delete the latest file, but I have to delete the rest.
Please advice me how can we achieve this via shell script. Your response is highly appreciated.
There are many ways to do this. The two primary approaches are either to (1) use the actual file date to determine whether the files are removed or (2) use the date embedded in the filename to determine the file date. Both have advantages and pitfalls. What you seem to be asking is to remove files 60 days older than the latest date embedded in the filename or 2.
As you have indicated, you may have a number of files with dates mixed relatively close to the end and you may need to adjust the date. Rather than just having the script parse for a maximum file date string contained in the file, you can prompt for the end date to measure 60 days back from. Otherwise, just scan each embedded date and find the max, and subtract 60 days from there. The following script prompts for an end_date.
In fact, the following script contains code to remove files by both methods (and sample data). The code to remove based on the actual file create date ( (1) above ) is commented out below the code that uses the embedded date. Look over the script and understand what it does. It is fairly well commented. NOTE the actual rm command is commented out to prevent accidents (even though it requires you to enter YES to confirm removal). Uncomment the rm line to be able to actually remove files. Drop a comment if you have questions:
#!/bin/bash
oifs="$IFS" # save current IFS (internal field separator) (default ' \t\n')
IFS=$'\n' # set IFS to only break on space
## prompt for path containing files & read
printf "\n enter the path to files to remove (no ending '/'): "
read -r rmpath
## validate directory
[ -d "$rmpath" ] || { printf "\nerror: bad path '%s'\n\n" "$rmpath"; exit 1; }
## prompt for ending date of files to keep
printf "\n enter the _end_ date of files to keep 'yyyymmdd' : "
read -r enddatestr
IFS="$oifs" # reset IFS to original
enddt=$(date -d "$enddatestr" +%s) # get enddt in seconds since epoch
enddt=$((enddt - (60 * 24 * 3600))) # subtract 60 days
declare -a rmarray
## Using embedded filename date
mdate=$(date -d "#$enddt" +%Y%m%d) # get mdate string to compare to filename
## fill rmarray with file dates older than mdate
for i in $(find "$rmpath" -maxdepth 1 -type f); do
ffname="${i##*/}" # full filename component
fname=${ffname%.*} # filename w/o extension
fdate="${fname##*_}" # get file date string
## if fdate before mdate, add to remove array
[ "$mdate" -gt "$fdate" ] && rmarray+=( "$i" )
done
# ### Using actual file creation date
# tgtfile=/tmp/tgt_$(date +%s) # tmp filename to measure against
#
# ## create temp file to measure against with find & set trap to remove
# touch -t $(date -d "#${enddt}" +%Y%m%d%H%M.%S) "$tgtfile" &&
# trap 'rm -rf "$tgtfile"' 0
#
# ## fill array with filenames to remove
# rmarray=( $(find "$rmpath" -maxdepth 1 -type f ! -newer $tgtfile) )
## verify files are contained in rmarray
[ "${#rmarray[#]}" -lt 1 ] && {
printf "\n No files matched the dates for removal.\n\n"
exit 1
}
## print files that will be removed
printf "\n ** the following files will be removed **\n\n"
for i in "${rmarray[#]}"; do
ls -al "$i"
done
## prompt for actual removal
printf "\n Continue with ACTUAL removal (YES to remove) : "
read ans
if [ "$ans" = "YES" ]; then
for i in "${rmarray[#]}"; do
# rm "$i" # NOTE: 'rm' is commented, uncomment to really delete
done
else
printf "\n You entered '%s' (not YES), no removal performed.\n\n" "$ans"
fi
exit 0
test directory:
$ls -l dat/fstst
total 0
-rw-r--r-- 1 david david 0 Nov 27 01:10 vtm_data_12month_20140301.control
-rw-r--r-- 1 david david 0 Nov 27 01:10 vtm_data_12month_20140301.txt
-rw-r--r-- 1 david david 0 Nov 27 01:10 vtm_mtd_20130622.control
-rw-r--r-- 1 david david 0 Nov 27 01:10 vtm_mtd_20130622.txt
-rw-r--r-- 1 david david 0 Nov 27 01:10 vtm_ytd_20131031.control
-rw-r--r-- 1 david david 0 Nov 27 01:10 vtm_ytd_20131031.txt
use:
$ bash rmfiles_60days.sh
enter the path to files to remove (no ending '/'): dat/fstst
enter the _end_ date of files to keep 'yyyymmdd' : 20140301
** the following files will be removed **
-rw-r--r-- 1 david david 0 Nov 27 01:10 dat/fstst/vtm_mtd_20130622.txt
-rw-r--r-- 1 david david 0 Nov 27 01:10 dat/fstst/vtm_ytd_20131031.control
-rw-r--r-- 1 david david 0 Nov 27 01:10 dat/fstst/vtm_ytd_20131031.txt
-rw-r--r-- 1 david david 0 Nov 27 01:10 dat/fstst/vtm_mtd_20130622.control
Continue with ACTUAL removal (YES to remove) : YES
result:
$ ls -l dat/fstst
total 0
-rw-r--r-- 1 david david 0 Nov 27 01:10 vtm_data_12month_20140301.control
-rw-r--r-- 1 david david 0 Nov 27 01:10 vtm_data_12month_20140301.txt
The following is an example using the actual file date:
test directory:
$ls -l dat/tst
total 324
-rw-r--r-- 1 david david 74 Sep 9 01:23 1.txt
-rw-r--r-- 1 david david 74 Sep 9 01:23 2.txt
-rw-r--r-- 1 david david 201 Aug 1 03:47 3line.dat
-rw-r--r-- 1 david david 205 Aug 1 03:35 3line.dat.sav
-rw-r--r-- 1 david david 88 Aug 13 04:05 catfile.txt
-rw-r--r-- 1 david david 39 Jul 4 14:40 comma
-rw-r--r-- 1 david david 291 Sep 23 03:00 createfile.txt
-rw-r--r-- 1 david david 11 Jul 17 03:54 data.dat
-rw-r--r-- 1 david david 8 Jul 17 03:54 datb.dat
-rw-r--r-- 1 david david 369 Oct 2 14:25 dia.txt
-rw-r--r-- 1 david david 36 Nov 6 15:51 dicta.dat
-rw-r--r-- 1 david david 23895 Sep 9 17:14 dna.dat
-rw-r--r-- 1 david david 243 Nov 4 23:07 domain.dat
-rw-r--r-- 1 david david 276 Nov 23 00:32 ecread.dat
(snip)
use:
$ bash rmfiles_60days.sh
enter the path to files to remove (no ending '/'): dat/tst
enter the _end_ date of files to keep 'yyyymmdd' : 20141031
** the following files will be removed **
-rw-r--r-- 1 david david 205 Aug 1 03:35 dat/tst/3line.dat.sav
-rw-r--r-- 1 david david 29 Jun 29 02:23 dat/tst/f1f2.dat
-rw-r--r-- 1 david david 8 Jul 17 03:54 dat/tst/datb.dat
-rw-r--r-- 1 david david 60 Jul 27 23:24 dat/tst/vowels.txt
-rw-r--r-- 1 david david 134 Aug 11 00:32 dat/tst/outfile.txt
-rw-r--r-- 1 david david 4622 Jun 26 02:49 dat/tst/single.xml
-rw-r--r-- 1 david david 99 Jul 4 14:51 dat/tst/hostnm
-rw-r--r-- 1 david david 115 Aug 7 01:35 dat/tst/ltags.txt
-rw-r--r-- 1 david david 122 Aug 29 11:11 dat/tst/hh.dat
-rw-r--r-- 1 david david 509 Jul 21 17:28 dat/tst/orders.txt
-rw-r--r-- 1 david david 205 Jun 27 01:06 dat/tst/table.html
(snip)
Continue with ACTUAL removal (YES to remove) : YES
result:
$ ls -l dat/tst
total 168
-rw-r--r-- 1 david david 74 Sep 9 01:23 1.txt
-rw-r--r-- 1 david david 74 Sep 9 01:23 2.txt
-rw-r--r-- 1 david david 291 Sep 23 03:00 createfile.txt
-rw-r--r-- 1 david david 369 Oct 2 14:25 dia.txt
-rw-r--r-- 1 david david 36 Nov 6 15:51 dicta.dat
-rw-r--r-- 1 david david 23895 Sep 9 17:14 dna.dat
-rw-r--r-- 1 david david 243 Nov 4 23:07 domain.dat
-rw-r--r-- 1 david david 276 Nov 23 00:32 ecread.dat
-rw-r--r-- 1 david david 93 Nov 2 21:43 empdata.dat
(snip)

Range with leading zero in bash

How to add leading zero to bash range?
For example, I need cycle 01,02,03,..,29,30
How can I implement this using bash?
In recent versions of bash you can do:
echo {01..30}
Output:
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Or if it should be comma separated:
echo {01..30} | tr ' ' ','
Which can also be accomplished with parameter expansion:
a=$(echo {01..30})
echo ${a// /,}
Output:
01,02,03,04,05,06,07,08,09,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30
another seq trick will work:
seq -w 30
if you check the man page, you will see the -w option is exactly for your requirement:
-w, --equal-width
equalize width by padding with leading zeroes
You can use seq's format option:
seq -f "%02g" 30
A "pure bash" way would be something like this:
echo {0..2}{0..9}
This will give you the following:
00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
Removing the first 00 and adding the last 30 is not too hard!
This works:
printf " %02d" $(seq 1 30)

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