I have a list of more than 1000 files on the following format.
0521865417_roman_pottery_in_the_archaeological_record_2007.pdf
0521865476_power_politics_and_religion_in_timurid_iran_2007.pdf
0521865514_toward_a_theory_of_human_rights_religion_law_courts_2006.pdf
0521865522_i_was_wrong_the_meanings_of_apologies_2008.pdf
I am on Linux and want to change them as follows
2007_roman_pottery_in_the_archaeological_record.pdf
2007_power_politics_and_religion_in_timurid_iran.pdf
2006_toward_a_theory_of_human_rights_religion_law_courts.pdf
2008_i_was_wrong_the_meanings_of_apologies.pdf
Using rename and awk I managed to get
2007_roman_pottery_in_the_archaeological_record_2007.pdf
2007_power_politics_and_religion_in_timurid_iran_2007.pdf
2006_toward_a_theory_of_human_rights_religion_law_courts_2006.pdf
2008_i_was_wrong_the_meanings_of_apologies_2008.pdf
The remaining task is now to remove the last field that holds the year.
A solution that uses sed to generate the new names and the rename commands then pipes them to bash:
ls -1 | sed -r 's/[0-9]*_([A-Za-z_]*)_[a-z]{3}_([0-9]{4})\.pdf$/mv & \2_\1.pdf/g' | bash
A work around from where you left of...
echo 2007_roman_pottery_in_the_archaeological_record_2007.pdf | awk -F '_' '{$NF=""; OFS="_"; print substr($0, 0, length($0)-1)".pdf";}'
Related
I'm new to batch. I want iterate through a list and use the output content to replace a string in another file.
ls -l somefile | grep .txt | awk 'print $4}' | while read file
do
toreplace="/Team/$file"
sed 's/dataFile/"$toreplace"/$file/ file2 > /tmp/test.txt
done
When I run the code I get the error
sed: 1: "s/dataFile/"$torepla ...": bad flag in substitute command: '$'
Example of somefile with which has list of files paths
foo/name/xxx/2020-01-01.txt
foo/name/xxx/2020-01-02.txt
foo/name/xxx/2020-01-03.txt
However, my desired output is to use the list of file paths in somefile directory to replace a string in another file2 content. Something like this:
This is the directory of locations where data from /Team/foo/name/xxx/2020-01-01.txt ............
I'm not sure if I understand your desired outcome, but hopefully this will help you to figure out your problem:
You have three files in a directory:
TEAM/foo/name/xxx/2020-01-02.txt
TEAM/foo/name/xxx/2020-01-03.txt
TEAM/foo/name/xxx/2020-01-01.txt
And you have another file called to_be_changed.txt which contains the text This is the directory of locations where data from TO_BE_REPLACED ............ and you want to grab the filenames of your three files and insert them into your to_be_changed.txt file, you can do it with:
while read file
do
filename="$file"
sed "s/TO_BE_REPLACED/${filename##*/}/g" to_be_changed.txt >> changed.txt
done < <(find ./TEAM/ -name "*.txt")
And you will then have made a file called changed.txt which contains:
This is the directory of locations where data from 2020-01-02.txt ............
This is the directory of locations where data from 2020-01-03.txt ............
This is the directory of locations where data from 2020-01-01.txt ............
Is this what you're trying to achieve? If you need further clarification I'm happy to edit this answer to provide more details/explanation.
ls -l somefile | grep .txt | awk 'print $4}' | while read file
No. No, no, nono.
ls -l somefile is only going to show somefile unless it's a directory.
(Don't name a directory "somefile".)
If you mean somefile.txt, please clarify in your post.
grep .txt is going to look through the lines presented for the three characters txt preceded by any character (the dot is a regex wildcard). Since you asked for a long listing of somefile it shouldn't find any, so nothing should be passed along.
awk 'print $4}' is a typo which won't compile. awk will crash.
Keep it simple. What I suspect you meant was
for file in *.txt
Then in
toreplace="/Team/$file"
sed 's/dataFile/"$toreplace"/$file/ file2 > /tmp/test.txt
it's unlear what you expect $file to be - awk's $4 from an ls -l seems unlikely.
Assuming it's the filenames from the for above, then try
sed "s,dataFile,/Team/$file," file2 > /tmp/test.txt
Does that help? Correct me as needed. Sorry if I seem harsh.
Welcome to SO. ;)
I am using Ubuntu Linux and grepping info out of a file (lets say filename.log) and want to save the file using some of the info inside of (filename.log).
example:
The info in the (filename.log) has version_name and date.
When displaying this info on screen using cat it will display:
version_name=NAME
date=TODAY
I then want to save the file as NAME-TODAY.log and have no idea how to do this.
Any help will be appreciated
You can chain a bunch of basic linux commands with the pipe character |. Combined with a thing called command substitution (taking the output of a complex command, to use in another command. syntax: $(your command)) you can achieve what you want to do.
This is what I came up with, based on your question:
cp filename.log $(grep -E "(version_name=)|(date=)" filename.log | cut -f 2 -d = | tr '\n' '-' | rev | cut -c 2- | rev).log
So here I used cp, $(), grep, cut, tr and finally rev.
Since you said you had no idea where to start, let me walk you trough this oneliner:
cp - it is used to copy the filename.log file to a new file,
with the name based on the values of version_name and date (step 2 and up)
command substitution $() the entire command between the round brackets is 'resolved' before finishing the cp command in step 1. e.g. in your example it would be NAME-TODAY. notice the .log at the end outside of the round brackets to give it a proper file extension. The output of this command in your example will be NAME-TODAY.log
grep -E "(version_name=)|(date=)" grep with regexp flag -E to be able to do what we are doing. Matches any lines that contain version_name= OR date=. The expected output is:
version_name=NAME
date=TODAY
cut -f 2 -d = because I am not interested in version_name
, but instead in the value associated with that field, I use cut to split the line at the equals character = with the flag -d =. I then select the value behind the equals character (the second field) with the flag -f 2. The expected output is:
NAME
TODAY
tr '\n' '-' because grep outputs on multiple lines, I want to remove all new lines and replace them with a dash. Expected output:
NAME-TODAY-
rev | cut -c 2- | rev I am grouping these. rev reverses the word I have created. with cut -c 2- I cut away all characters starting from the second character of the reversed word. This is required because I replaced new lines with dashes and this means I now have NAME-TODAY-. Basicly this is just an extra step to remove the last dash. See expected outputs of each step:
-YADOT-EMAN
YADOT-EMAN
NAME-TODAY
remember this value is in the command substituion of step 2, so the end result will be:
cp filename.log NAME-TODAY.log
I manged to solve this by doing the following: grep filename.log > /tmp/file.info && filename=$(echo $(grep "version_name" /tmp/filename.info | cut -d " " -f 3)-$(grep "date" /tmp/filename.info | cut -d " " -f 3)-$filename.log
I have some files say about 1000 numbers.. Wanted to rename those files in such a way that, wanted to cut out only few chars from file name and copy it to some other directory.
Ex: Original file name.
vfcon062562~19.xml
vfcon058794~29.xml
vfcon072009~3.xml
vfcon071992~10.xml
vfcon071986~2.xml
vfcon071339~4.xml
vfcon069979~43.xml
Required O/P is cutting the ~and following chars.
O/P Ex:
vfcon058794.xml
vfcon062562.xml
vfcon069979.xml
vfcon071339.xml
vfcon071986.xml
vfcon071992.xml
vfcon072009.xml
But want to place n different directory.
If you are using bash or similar you can use the following simple loop:
for input in vfcon*xml
do
mv $input targetDir/$(echo $input | awk -F~ '{print $1".xml"}')
done
Or in a single line:
for input in vfcon*xml; do mv $input targetDir/$(echo $input | awk -F~ '{print $1".xml"}'); done
This uses awk to separate everything before ~ using it as a field separator and printing the first column and appending ".xml" to create the output file name. All this is prepended with the targetDir which can be a full path.
If you are using csh / tcsh then the syntax of the loop will be slightly different but the commands will be the same.
I like to make sure that my data set is correct prior to changing anything so I would put that into a variable first and then check over it.
files=$(ls vfcon*xml)
echo $files | less
Then, like #Stefan said, use a loop:
for i in $files
do
mv "$i" "$( echo "$file" | sed 's/~[0-9].//g')"
done
If you need help with bash you can use http://www.shellcheck.net/
Here is my problem,
I have a folder where is stored multiple files with a specific format:
Name_of_file.TypeMM-DD-YYYY-HH:MM
where MM-DD-YYYY-HH:MM is the time of its creation. There could be multiple files with the same name but not the same time of course.
What i want is a script that can keep the 3 newest version of each file.
So, I found one example there:
Deleting oldest files with shell
But I don't want to delete a number of files but to keep a certain number of newer files. Is there a way to get that find command, parse in the Name_of_file and keep the 3 newest???
Here is the code I've tried yet, but it's not exactly what I need.
find /the/folder -type f -name 'Name_of_file.Type*' -mtime +3 -delete
Thanks for help!
So i decided to add my final solution in case anyone liked to get it. It's a combination of the 2 solutions given.
ls -r | grep -P "(.+)\d{4}-\d{2}-\d{2}-\d{2}:\d{2}" | awk 'NR > 3' | xargs rm
One line, super efficiant. If anything changes on the pattern of date or name just change the grep -P pattern to match it. This way you are sure that only the files fitting this pattern will get deleted.
Can you be extra, extra sure that the timestamp on the file is the exact same timestamp on the file name? If they're off a bit, do you care?
The ls command can sort files by timestamp order. You could do something like this:
$ ls -t | awk 'NR > 3' | xargs rm
THe ls -t lists the files by modification time where the newest are first.
The `awk 'NR > 3' prints out the list of files except for the first three lines which are the three newest.
The xargs rm will remove the files that are older than the first three.
Now, this isn't the exact solution. There are possible problems with xargs because file names might contain weird characters or whitespace. If you can guarantee that's not the case, this should be okay.
Also, you probably want to group the files by name, and keep the last three. Hmm...
ls | sed 's/MM-DD-YYYY-HH:MM*$//' | sort -u | while read file
do
ls -t $file* | awk 'NR > 3' | xargs rm
done
The ls will list all of the files in the directory. The sed 's/\MM-DD-YYYY-HH:MM//' will remove the date time stamp from the files. Thesort -u` will make sure you only have the unique file names. Thus
file1.txt-01-12-1950
file2.txt-02-12-1978
file2.txt-03-12-1991
Will be reduced to just:
file1.txt
file2.txt
These are placed through the loop, and the ls $file* will list all of the files that start with the file name and suffix, but will pipe that to awk which will strip out the newest three, and pipe that to xargs rm that will delete all but the newest three.
Assuming we're using the date in the filename to date the archive file, and that is possible to change the date format to YYYY-MM-DD-HH:MM (as established in comments above), here's a quick and dirty shell script to keep the newest 3 versions of each file within the present working directory:
#!/bin/bash
KEEP=3 # number of versions to keep
while read FNAME; do
NODATE=${FNAME:0:-16} # get filename without the date (remove last 16 chars)
if [ "$NODATE" != "$LASTSEEN" ]; then # new file found
FOUND=1; LASTSEEN="$NODATE"
else # same file, different date
let FOUND="FOUND + 1"
if [ $FOUND -gt $KEEP ]; then
echo "- Deleting older file: $FNAME"
rm "$FNAME"
fi
fi
done < <(\ls -r | grep -P "(.+)\d{4}-\d{2}-\d{2}-\d{2}:\d{2}")
Example run:
[me#home]$ ls
another_file.txt2011-02-11-08:05
another_file.txt2012-12-09-23:13
delete_old.sh
not_an_archive.jpg
some_file.exe2011-12-12-12:11
some_file.exe2012-01-11-23:11
some_file.exe2012-12-10-00:11
some_file.exe2013-03-01-23:11
some_file.exe2013-03-01-23:12
[me#home]$ ./delete_old.sh
- Deleting older file: some_file.exe2012-01-11-23:11
- Deleting older file: some_file.exe2011-12-12-12:11
[me#home]$ ls
another_file.txt2011-02-11-08:05
another_file.txt2012-12-09-23:13
delete_old.sh
not_an_archive.jpg
some_file.exe2012-12-10-00:11
some_file.exe2013-03-01-23:11
some_file.exe2013-03-01-23:12
Essentially, but changing the file name to dates in the form to YYYY-MM-DD-HH:MM, a normal string sort (such as that done by ls) will automatically group similar files together sorted by date-time.
The ls -r on the last line simply lists all files within the current working directly print the results in reverse order so newer archive files appear first.
We pass the output through grep to extract only files that are in the correct format.
The output of that command combination is then looped through (see the while loop) and we can simply start deleting after 3 occurrences of the same filename (minus the date portion).
This pipeline will get you the 3 newest files (by modification time) in the current dir
stat -c $'%Y\t%n' file* | sort -n | tail -3 | cut -f 2-
To get all but the 3 newest:
stat -c $'%Y\t%n' file* | sort -rn | tail -n +4 | cut -f 2-
File1:
<a>hello</b> <c>foo</d>
<a>world</b> <c>bar</d>
Is an example of the file this would work on. How can one remove all strings which have a <c>*</d> using sed?
The following line will remove all text from <c> to </d> inclusive:
sed -e 's/<c>.*<\/d>//'
The bit inside the s/...// is a regular expression, not really a wildcard in the same way as the shell uses, so anything you can put in a regular expression you can put in there.
if all your data is like that of the example
# gawk 'BEGIN{FS=" <c>"}{print $1}' file
<a>hello</b>
<a>world</b>
Great Swiss-Army knife!
I modified it to pull header info out of eMails for an archiving script. It involved renaming the IMAP eMails with both date and sender info (otherwise IMAP just numbered 1, 2, 3, etc.). Here's the two mods:
for i in $mailarray; do date -d $(less -f $i | grep -im 1 "Date:\ " | sed -e 's_^.*\(ate: \)__') +%F_%T%Z; done
for i in $mailarray; do less -f "$i" | grep -iEm 1 "From:\ " | sed -e 's_^.*\(rom\).*<\|^.*\(rom:\).__' | sed -e 's_#.*$__'; done
They saved a great deal of extraneous coding. Thank you.