I'm writting some zsh functions using the powerful completion feature. The computation of my completions take some times and I want to make use of the completion caching policy. From the zsh manual (https://zsh.sourceforge.io/Doc/Release/Completion-System.html) I found this code snippet
example_caching_policy () {
# rebuild if cache is more than a week old
local -a oldp
oldp=( "$1"(Nm+7) )
(( $#oldp ))
}
I couldn't find any explanation on the (Nm+7) syntax, what does Nm means ? With try and error I could find out that for example Nms+1 would change the cache policy to 1 second, while Nmh+1 to 1 hour. But where can I find the general (NmX+N) construct explanation ?
Same what does exactly means the line (( $#oldp )) ?
I can explain the (Nm+7)
man zshexpn, search for Glob Qualifiers
a[Mwhms][-|+]n
files accessed exactly n days ago. Files accessed within the last n days are selected using a negative
value for n (-n). Files accessed more than n days ago are selected by a positive n value (+n). Op‐
tional unit specifiers `M', `w', `h', `m' or `s' (e.g. `ah5') cause the check to be performed with
months (of 30 days), weeks, hours, minutes or seconds instead of days, respectively. An explicit `d'
for days is also allowed.
Any fractional part of the difference between the access time and the current part in the appropriate
units is ignored in the comparison. For instance, `echo *(ah-5)' would echo files accessed within the
last five hours, while `echo *(ah+5)' would echo files accessed at least six hours ago, as times
strictly between five and six hours are treated as five hours.
m[Mwhms][-|+]n
like the file access qualifier, except that it uses the file modification time.
N stands for NULL_GLOB, if zsh matches nothing, it will remove the pattern.
Without this N option, if it matches nothing it will print an error.
Example with 4 files
$ touch lal # = updates file modification date to now
$ ls
lal lil lol tut
$ ls l*(m+7)
lil lol
# files older than 7 days starting with l
$ ls l*(m-7)
lal
# files younger than 7 days starting with l
$ ls l*(m+200)
zsh: no match
# no files older than 200 days
$ ls l*(Nm+200)
lal lil lol tut
# N = NULL_GLOB made disappear the non-matching pattern so it's just ls
I downloaded all the pictures from an old spanish social network and the Last modified date is the date that I downloaded the file instead of the real date. I would like to order all the pictures by Last modified date to upload to google photos using the correct date.
I know the real date because the image name has the date and time and I would make a script to change de last modified date using the name of the file. I don't care about the time.
Name of the file format: "yyyy-mm-dd hh-mm-ss.jpg"
I'm using windows, but is not a problem to install Unix to execute some bash script, because it may probabbly be easier.
Is there any way to do this?
Thanks in advance
You can use powershell $(Get-Item *Filename*).creationtime=$(Get-Date "mm/dd/yyyy hh:mm am/pm") in a batch file or command prompt to change the creation date of a file. for more information look at this. In your case you could do:
for %%i in (<folder with all the files path>) do (
set name=%%~i
set name=%name:~0,19%
powershell $(Get-Item %%~i).creationtime=$(Get-Date "%name:~5,2%/%name:~8,2%/%name:~0,4% %name:~11,2%:%name:~14,2%")
)
Note: I have not tested this since I do not have the data set for it.
First of all, I'm very new to programming and so would need your help in writing a perl script to do the following on windows.
I have a big log file with timestamp (1gb) and its difficult to read the logs as it takes a lot of time to open. so my requirement is to copy the logs from the bigger log file for the last one hour and paste it to another file and then copy the next 1 hr of data to different file(so we will have 24 files for a day). The next day the data in these files needs to be over written or delete & create a new file.
Sample log :
09092016-00:02:00,..................
09092016-00:02:08,..................
09092016-00:02:15,..................
09092016-00:02:18,..................
Please help me with this and thanks for your help in advance.
Thanks,
A simpler solution would be to use the split command to split the files into manageable sizes.
split -l 1000 logfile
Will split your logfile into smaller files of 1000 lines each.
You can then just use grep to find the files that contain the day you need.
grep 09092016 logfile*
for example:
logfile="./log"
while read -r d m y h; do
grep "^$d$m$y-$h" "$logfile" > "partial-${y}${m}{$d}-${h}.log"
done < <(sed -n 's/\(..\)\(..\)\(....\)-\(..\)\(.*\)/\1 \2 \3 \4/p' "$logfile" | sort -u)
easy, but not efficient. It reads the whole big logfile 25x for the split. (1x for gathering the existing ddmmyyyy-hh lines in the log, and again for every different found date-hour.)
I am looking to compare the last updated date and time of a file against the current date.
To date the date/time of the file I am splitting up the time and date tag like this:
find . -maxdepth 1 -type f -name \*.dat -printf "%Tm,%Td,%TH,%TM\n"
And I am looking to match this against the current date:
date '+%m,%d,%H,%M'
Both are in the same format - DD MM HH MM - and I would like to compare if the time is more than two minutes out.
What would be my best option for coding this - can I use the values I have obtained?
Or even - would there be an easier method?
Yes, there is an easier way.
You can use
find -mmin -2
for finding all the entries which have been changed less than two minutes ago.
Or, of course, use
find -mmin +2
to find the entries which are older than two minutes (have been changed more than two minutes ago).
Project moving forwards, I can see why creating .bat files to do things can become addictive!
I can now save somefile.txt at regular intervals, I then rename somefile.txt by adding the time and date to create a unique file name
ren somefile.txt somefile_%time:~0,2%%time:~3,2%-%date:~-10,2%%date:~3,2%%date:~-4,4%.txt
As an example, the code above has just renamed somefile.txt to somefile_1317_13022011.txt (1317hrs on 13th February 2011)
I ran
ren somefile.txt somefile_%time:~0,2%%time:~3,2%-%date:~-10,2%%date:~7,2%%date:~-4,4%.txt
yesterday, it ran successfully until midnight, and then it crashed (syntax error) although it was saving as 12012011 for the date (12th Jan 2011) instead of the correct date of 12022011.
Will the current version ran past midnight? Am I confusing myself with UK vs US date format?
Animuson gives a decent way to do it, but no help on understanding it. I kept looking and came across a forum thread with this commands:
Echo Off
IF Not EXIST n:\dbfs\doekasp.txt GOTO DoNothing
copy n:\dbfs\doekasp.txt n:\history\doekasp.txt
Rem rename command is done twice (2) to allow for 1 or 2 digit hour,
Rem If before 10am (1digit) hour Rename starting at location (0) for (2) chars,
Rem will error out, as location (0) will have a space
Rem and space is invalid character for file name,
Rem so second remame will be used.
Rem
Rem if equal 10am or later (2 digit hour) then first remame will work and second will not
Rem as doekasp.txt will not be found (remamed)
ren n:\history\doekasp.txt doekasp-%date:~4,2%-%date:~7,2%-%date:~10,4%_#_%time:~0,2%h%time:~3,2%m%time:~6,2%s%.txt
ren n:\history\doekasp.txt doekasp-%date:~4,2%-%date:~7,2%-%date:~10,4%_#_%time:~1,1%h%time:~3,2%m%time:~6,2%s%.txt
I always name year first YYYYMMDD, but wanted to add time. Here you will see that he has given a reason why 0,2 will not work and 1,1 will, because (space) is an invalid character. This opened my eyes to the issue. Also, by default you're in 24hr mode.
I ended up with:
ren Logs.txt Logs-%date:~10,4%%date:~7,2%%date:~4,2%_%time:~0,2%%time:~3,2%.txt
ren Logs.txt Logs-%date:~10,4%%date:~7,2%%date:~4,2%_%time:~1,1%%time:~3,2%.txt
Output:
Logs-20121707_1019
Digging up the old thread because all solutions have missed the simplest fix...
It is failing because the substitution of the time variable results in a space in the filename, meaning it treats the last part of the filename as a parameter into the command.
The simplest solution is to just surround the desired filename in quotes "filename".
Then you can have any date pattern you want (with the exception of those illegal characters such as /,\,...)
I would suggest reverse date order YYYYMMDD-HHMM:
ren "somefile.txt" "somefile-%date:~10,4%%date:~7,2%%date:~4,2%-%time:~0,2%%time:~3,2%.txt"
following should be your right solution
ren somefile.txt somefile_%time:~0,2%%time:~3,2%-%DATE:/=%.txt
I took the above but had to add one more piece because it was putting a space after the hour which gave a syntax error with the rename command.
I used:
set HR=%time:~0,2%
set HR=%Hr: =0%
set HR=%HR: =%
rename c:\ops\logs\copyinvoices.log copyinvoices_results_%date:~10,4%-%date:~4,2%-%date:~7,2%_%HR%%time:~3,2%.log
This gave me my format I needed:
copyinvoices_results_2013-09-13_0845.log
problem in %time:~0,2% can't set to 24 hrs format, ended with space(1-9), instead of 0(1-9)
go around with:
set HR=%time:~0,2%
set HR=%Hr: =0% (replace space with 0 if any <has a space in between : =0>)
then replace %time:~0,2% with %HR%
good luck
ls | xargs -I % mv % %_`date +%d%b%Y`
One line is enough.
ls all files/dirs under current dir and append date to each file.
I tried to do the same:
<fileName>.<ext> --> <fileName>_<date>_<time>.<ext>
I found that :
rename 's/(\w+)(\.\w+)/$1'$(date +"%Y%m%d_%H%M%S)'$2/' *