Change file extension of several files in one command - bash

I'm using a binary to convert a file type to another:
./rcc -binary themes/redTheme.qrc -o redTheme.rcc
I have several .qrc files and would like to convert them as well, but in the same command. How to regroup them rename and their type? I'm struggling to find a way to keep the file name.
I was thinking about binding my rcc binary to a pipe that would execute a loop converting the .qrc files.

I hope you can bind this with your binary:
ls *.qrc | xargs -I {} sh -c 'mv $1 `basename $1 .html`.rcc' - {}

You could use something like this:
$ for i in *.qrc ; do echo mv -v ${i} ${i%.qrc}.new ; done
See here an example where I have a couple of files with qrc ending and I mv them to .new
$ l
total 0
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 1.qrc
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 10.qrc
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 2.qrc
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 3.qrc
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 4.qrc
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 5.qrc
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 6.qrc
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 7.qrc
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 8.qrc
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 9.qrc
$ for i in *.qrc ; do mv -v ${i} ${i%.qrc}.new ; done
1.qrc -> 1.new
10.qrc -> 10.new
2.qrc -> 2.new
3.qrc -> 3.new
4.qrc -> 4.new
5.qrc -> 5.new
6.qrc -> 6.new
7.qrc -> 7.new
8.qrc -> 8.new
9.qrc -> 9.new
$ l
total 0
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 1.new
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 10.new
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 2.new
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 3.new
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 4.new
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 5.new
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 6.new
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 7.new
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 8.new
-rw-r--r-- 1 chl wheel 0 Mar 9 16:22 9.new
$
Of course you have to use your rcc tool instead of mv and use your ending instead of .new

Related

How to touch files with different names and different date created

I am trying to create files with different date created:
$ touch -t 20{11..15}01120000 file_{1..5}.txt
$ ls -al
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 201201120000
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 201301120000
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 201401120000
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 201501120000
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 file_1.txt
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 file_2.txt
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 file_3.txt
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 file_4.txt
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 file_5.txt
As you can see all file's dates created are 2011 Jan 12 0.
How can I create files with a different year?
# this is what I want
-rw-r--r-- 1 shinokada staff 0 Jan 12 2011 file_1.txt
-rw-r--r-- 1 shinokada staff 0 Jan 12 2012 file_2.txt
-rw-r--r-- 1 shinokada staff 0 Jan 12 2013 file_3.txt
-rw-r--r-- 1 shinokada staff 0 Jan 12 2014 file_4.txt
-rw-r--r-- 1 shinokada staff 0 Jan 12 2015 file_5.txt
What is the best way?
touch command allows you to enter one one timestamp using -t option.
A traditional for-loop would be better:
for i in {1..5}; do touch -t 201${i}01120000 file_$i.txt; done
Shortly, but keeping filenumbers and years as separated variables
I think this is more readable, but...
Care about timezone!
filenum=1
for year in {2011..2015};do
TZ=UTC touch -t ${year}12312345 file-$((filenum++))
done
Then if you look about this, using a different timezone:
(Note that created date was Dec 31, 23h45')
TZ=UTC-1 ls -ltr
-rw-r--r-- 1 user user 0 jan 1 2012 file-1
-rw-r--r-- 1 user user 0 jan 1 2013 file-2
-rw-r--r-- 1 user user 0 jan 1 2014 file-3
-rw-r--r-- 1 user user 0 jan 1 2015 file-4
-rw-r--r-- 1 user user 0 jan 1 2016 file-5
Could you please try following, I would go with following approach with a for loop. Where I am providing year, number of files which needed, output file's initial value and same time value for all output files so that we can manage it in for loop.
cat script.bash
year=2011
numberoffiles="10"
time="01120000"
outputfileInitials="file"
nameSequence="1"
for ((i = 1 ; i <= numberoffiles ; i++ ));
do
touch -t $year$time "${outputfileInitials}_$nameSequence.txt"
(( nameSequence = nameSequence + 1 ))
(( year = year + 1 ))
done

Clean ten days and older logs using Shell

[root#amp logs]# ls -l
total 0
-rw-r--r-- 1 root root 0 Nov 23 17:51 lb-quarzcenter.log
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-01
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-02
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-03
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-04
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-05
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-06
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-07
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-08
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-10
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-11
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-12
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-13
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-14
-rw-r--r-- 1 root root 0 Nov 23 17:51 lb-quarzcenter.log.2019-02-15
-rw-r--r-- 1 root root 0 Feb 27 17:26 lb-quarzcenter.log.2019-02-09
How do I match a string with year-month-date and delete 10 days ago file according to lb-quarzcenter.log.*?
now=$(date +%s) # express current time as seconds since 1970-01-01
(( ten_days_ago = now - 60*60*24*10 )) # subtract 864000 seconds (10 days) from that
date_minus_ten=$(date +%F --date="#$ten_days_ago") # express that number as a YYYY-MM-DD
for filename in lb-quartzcenter.log.* ; do # loop over all matching files
filedate="${filename/lb-quartzcenter.log./}" # remove the filename part before the timestamp
if [[ $filedate < $date_minus_ten ]] ; then # if filename remainder is lexicographically lower...
rm -f "$filename" # ... remove the file
fi
done

Can I change the modified date, just the year, of a file or all files?

Can I change the modified date, just the year, of a file or all files?
I have been looking here which lead me to touch.
$ ls -l *.txt
-rw-r--r-- 1 kevin.smith mkpasswd 3319 Nov 21 2017 adjectives.txt
-rw-r--r-- 1 kevin.smith mkpasswd 25562 Aug 11 2015 checklist.txt
-rwxr-xr-x 1 kevin.smith mkpasswd 11347 May 9 14:28 cw_text.txt
-rw-r--r-- 1 kevin.smith mkpasswd 9260 May 9 14:31 cw_text2.txt
-rw-r--r-- 1 kevin.smith mkpasswd 4786 May 9 14:38 cw_text3.txt
-rw-r--r-- 1 kevin.smith mkpasswd 390 Jun 25 2014 Delete_log.txt
-rw-r--r-- 1 kevin.smith mkpasswd 6891 Jul 27 2015 log.txt
-rw-r--r-- 1 kevin.smith mkpasswd 53828 Jan 17 2017 pin1.txt
-rw-r--r-- 1 kevin.smith mkpasswd 39412 Jan 17 2017 pip2.txt
-rw-r--r-- 1 kevin.smith mkpasswd 167 Dec 5 2015 romeo.txt
$ touch -t 2018* *.txt
Expected Output: Would have only the year changed to 2018
$ ls -l *.txt
-rw-r--r-- 1 kevin.smith mkpasswd 3319 Nov 21 2018 adjectives.txt
-rw-r--r-- 1 kevin.smith mkpasswd 25562 Aug 11 2018 checklist.txt
-rwxr-xr-x 1 kevin.smith mkpasswd 11347 May 9 14:28 cw_text.txt
-rw-r--r-- 1 kevin.smith mkpasswd 9260 May 9 14:31 cw_text2.txt
-rw-r--r-- 1 kevin.smith mkpasswd 4786 May 9 14:38 cw_text3.txt
-rw-r--r-- 1 kevin.smith mkpasswd 390 Jun 25 2018 Delete_log.txt
-rw-r--r-- 1 kevin.smith mkpasswd 6891 Jul 27 2018 log.txt
-rw-r--r-- 1 kevin.smith mkpasswd 53828 Jan 17 2018 pin1.txt
-rw-r--r-- 1 kevin.smith mkpasswd 39412 Jan 17 2018 pip2.txt
-rw-r--r-- 1 kevin.smith mkpasswd 167 Dec 5 2018 romeo.txt
ls -l file.txt; touch -t "$(date -d "#$(stat -c '%Y' file.txt)" "+2020%m%d%H%M")" file.txt; ls -l file.txt
# ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
-rw-r--r-- 1 jackman jackman 0 Oct 23 16:47 file.txt
-rw-r--r-- 1 jackman jackman 0 Oct 23 2020 file.txt
You'll need to use a for loop to iterate over the files, and query/update the mtime one-by-one

How can I rename of old files?

I have a directory like this:
-rw-r--r-- 1 root root 0 Jan 7 15:04 tmp_FILE2015_123_1_3123.LOG
-rw-r--r-- 1 root root 0 Jan 7 15:04 tmp_FILE2015_133_1_3123.LOG
-rw-r--r-- 1 root root 0 Jan 7 16:04 tmp_FILE2015_133_1_3125.LOG
-rw-r--r-- 1 root root 0 Jan 7 16:04 tmp_FILE2015_133_1__3223125.LOG
-rw-r--r-- 1 root root 0 Jan 7 16:04 tmp_FILE2015_133_1_3223125.LOG
I need to remove tmp_ and I can do like this:
for i in *; do s=$(sed -r 's/^(tmp_)(.*.LOG)/\2/' <<< $i); if [[ "$i" != "$s" ]]; then mv "$i" "$s"; fi; done;
But I need to do this for just older than 1 hour (modified time) files:
For example ( now: Jan 7 16:10 ):
-rw-r--r-- 1 root root 0 Jan 7 13:00 FILE2015_123_1_3123.LOG
-rw-r--r-- 1 root root 0 Jan 7 15:04 FILE2015_133_1_3123.LOG
-rw-r--r-- 1 root root 0 Jan 7 15:01 FILE2015_133_1_3125.LOG
-rw-r--r-- 1 root root 0 Jan 7 16:04 tmp_FILE2015_133_1__3223125.LOG
-rw-r--r-- 1 root root 0 Jan 7 16:10 tmp_FILE2015_133_1_3223125.LOG
How can I do that?
This will operate on all files modified in the last hour:
for orig_file in $(find . -type f -depth 1 -mtime -60m); do
new_file="${orig_file#./tmp_}"
if [[ "$new_file" != "$orig_file" ]]; then
mv "$orig_file" "$new_file";
fi;
done
If you want to operate on files older than one hour use +60m instead
I've changed the use of sed to use some built in bash functionality.

how to print 3 lines in 1 column?

I've a file like below.
ab13p29im-sss29511
0
Jan 12 22:43
ab13p29im-sss29531
0
Jan 12 22:43
ab13p29im-sss29512
0
Feb 2 16:11
ab13p29im-sss29522
0
Feb 2 16:12
ab13p29im-sss29532
0
Feb 2 16:12
ab21p30im-sss30511
0
Jan 12 22:43
ab21p30im-sss30531
0
Jan 12 22:43
ab21p30im-sss30512
0
Feb 2 16:13
ab21p30im-sss30522
3
Feb 2 16:12
i want to print this is below format.
ab13p29im-sss29511 0 Jan 12 22:43
ab21p30im-sss30522 0 Feb 2 16:12
ab21p30im-sss30531 0 Jan 12 22:43
I'm using the command paste - - - < inputfile.But if any of the value is null, the format is all messed up like below?
ab13p29im-sss29511 0 Jan 12 22:43
ab21p30im-sss30522 0 ab21p30im-sss30531
0 Jan 12 22:43 ab21p30im-sss30523.
Like if there's no date for any host or if any value is null, it breaks the 3,3,3 pattern.
You like some like this:
awk 'ORS=NR%3?" ":RS' file
ab13p29im-sss29511 0 Jan 12 22:43
ab13p29im-sss29531 0 Jan 12 22:43
ab13p29im-sss29512 0 Feb 2 16:11
ab13p29im-sss29522 0 Feb 2 16:12
ab13p29im-sss29532 0 Feb 2 16:12
ab21p30im-sss30511 0 Jan 12 22:43
ab21p30im-sss30531 0 Jan 12 22:43
ab21p30im-sss30512 0 Feb 2 16:13
ab21p30im-sss30522 3 Feb 2 16:12
sed 'N;N;s/\n/ /g' YourFile
Load 2 lines, remove new line before printing it then cycle
you could secude by putting a pattern check to initiate the cycle like /[a-b0-9]\{9\}-[a-b0-9]\{8\}/!d; before first N
By using awk command and explicit concatenation:
$ awk 'NR % 3 == 1 { lines=$0 ; next } { lines=lines" "$0 } NR % 3 == 0 { print lines ; lines="" }' file
ab13p29im-sss29511 0 Jan 12 22:43
ab13p29im-sss29531 0 Jan 12 22:43
ab13p29im-sss29512 0 Feb 2 16:11
ab13p29im-sss29522 0 Feb 2 16:12
ab13p29im-sss29532 0 Feb 2 16:12
ab21p30im-sss30511 0 Jan 12 22:43
ab21p30im-sss30531 0 Jan 12 22:43
ab21p30im-sss30512 0 Feb 2 16:13
ab21p30im-sss30522 3 Feb 2 16:12

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