I am preparing shell script if folder created date is equal to the current date/time then it need to call the another script .
My requirement is script need to check only the folder created date against the current date/time not to the files date/time inside the folder .
Thanks in advance
You can have a look at the stat command and the change time of the folder. The change time gives the last date when the metadata of the folder was changed. (stackexchange link). Timestamps of permission changes are also included in this timestamp. This is maybe not exactly what you need.
For the current time, you can use the date command. You can compare timestamps between the stat and the date command if you print the output in seconds since the epoch.
stat --format="%Z" /path/to/folder
date +%s
Suppose TestToday is your Directory Name
*Please replace TestToday with your directory name in below script
Test.sh
ls -lrt | grep ^d | grep TestToday | grep "`date +%b`" | awk -v var="`date +%d | bc`" '$7==var {print $NF}' > DIR_NAME
DIR_EXISTS=`cat DIR_NAME`
#echo $DIR_EXISTS
if [ "$DIR_EXISTS" == "TestToday" ];then
echo "Calling Shell Script Here."
else
echo "Directed not created or updated today."
fi
Related
Lets say i have multiple filesnames e.g. R014-20171109-1159.log.20171109_1159.
I want to create a shell script which creates for every given date a folder and moves the files matching the date to it.
Is this possible?
For the example a folder "20171109" should be created and has the file "R014-20171109-1159.log.20171109_1159" on it.
Thanks
This is a typical application of a for-loop in bash to iterate thru files.
At the same time, this solution utilizes GNU [ shell param substitution ].
for file in /path/to/files/*\.log\.*
do
foldername=${file#*-}
foldername=${foldername%%-*}
mkdir -p "${foldername}" # -p suppress errors if folder already exists
[ $? -eq 0 ] && mv "${file}" "${foldername}" # check last cmd status and move
done
Since you want to write a shell script, use commands. To get date, use cut cmd like ex:
cat 1.txt
R014-20171109-1159.log.20171109_1159
cat 1.txt | cut -d "-" -f2
Output
20171109
is your date and create folder. This way you can loop and create as many folders as you want
Its actually quite easy(my Bash syntax might be a bit off) -
for f in /path/to/your/files*; do
## Check if the glob gets expanded to existing files.
## If not, f here will be exactly the pattern above
## and the exists test will evaluate to false.
[ -e "$f" ] && echo $f > #grep the file name for "*.log."
#and extract 8 charecters after "*.log." .
#Next check if a folder exists already with the name of 8 charecters.
#If not { create}
#else just move the file to that folder path
break
done
Main idea is from this post link. Sorry for not providing the actual code as i havent worked anytime recently on Bash
Below commands can be put in script to achieve this,
Assign a variable with current date as below ( use --date='n day ago' option if need to have an older date).
if need to get it from File name itself, get files in a loop then use cut command to get the date string,
dirVar=$(date +%Y%m%d) --> for current day,
dirVar=$(date +%Y%m%d --date='1 day ago') --> for yesterday,
dirVar=$(echo $fileName | cut -c6-13) or
dirVar=$(echo $fileName | cut -d- -f2) --> to get from $fileName
Create directory with the variable value as below, (-p : create directory if doesn't exist.)
mkdir -p ${dirVar}
Move files to directory to the directory with below line,
mv *log.${dirVar}* ${dirVar}/
I want to get the list of files from a directory and group them in an array or variable based on unique time stamp ( ls -ltr month , day ) using bash. this time stamp is 2-3 columns in range.
Any suggestions?
This is a one-liner don't know if is exactly what you are asking for:
array=($(ls -ltr | awk -v x=9 '{print $x}'))
It will create an array with the output of ls -ltr of the files
To print contents of the array:
printf "%s\n" "${array[#]}"
But is also worth checking this "Why you shouldn't parse the output of ls(1)"
I searched a while and tried it by myself but unable to get this sorted so far. My folder looks below, 4 files
1.txt, 2.txt, 3.txt, 4.txt, 5.txt, 6.txt
I want to print file modified time and echo the time stamp in it
#!/bin/bash
thedate= `ls | xargs stat -s | grep -o "st_mtime=[0-9]*" | sed "s/st_mtime=//g"` #get file modified time
files= $(ls | grep -Ev '(5.txt|6.txt)$') #exclud 5 and 6 text file
for i in $thedate; do
echo $i >> $files
done
I want to insert each timestamp to each file. but having "ambiguous redirect" error. am I doing it incorrectly? Thanks
In this case, files is a "list" of files, so you probably want to add another loop to handle them one by one.
Your description is slightly confusing but, if your intent is to append the last modification date of each file to that file, you can do something like:
for fspec in [1-4].txt ; do
stat -c %y ${fspec} >>${fspec}
done
Note I've used stat -c %y to get the modification time such as 2017-02-09 12:21:22.848349503 +0800 - I'm not sure what variant of stat you're using but mine doesn't have a -s option. You can still use your option, you just have to ensure it's done on each file in turn, probably something like (in the for loop above):
stat -s ${fspec} | grep -o "st_mtime=[0-9]*" | sed "s/st_mtime=//g" >>${fspec}
You can not redirect the output to several files as in > $files.
To process several files you need something like:
#!/bin/bash
for f in ./[0-4].txt ; do
# get file modified time (in seconds)
thedate="$(stat --printf='%Y\n' "$f")"
echo "$thedate" >> "$f"
done
If you want a human readable time format change %Y by %y:
thedate="$(stat --printf='%y\n' "$f")"
I'm new to linux command.
Now I would like to extract only date value from listed file name and compare as a date value.
Example:
/underdirectory
20080206
20080207
bk_20080208
I want to listed all above directories and compare date, means this directory is greater or smaller than which is according to specified date.
If all of the listed directories will be date, it's ok for condition check.
code
foreach date_directory ( ls )
if ( "$date_directory" >= "$fdate" && "$date_directory" <= "$tdate") then
echo ${target_del}${date_directory} >> ${output}
endif
end
But if include some words such as "bk_20080228" or "bk_20080228_bk" or "20080228_bk" or "20080228_tt", there condition check will be error.
bk_20080208 want to take only 20080208.
please help me.
Thanks!
use grep regexp grep -Eo '[0-9]{4}[0-9]{2}[0-9]{2}' or grep -Eo '[0-9]+'
shopt -s nullglob
for i in *;do echo $i|sed 's/.*_//';done
the output can be redirected as per needs.
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-