Is there any way to use standard tools (not programming scripts) to parse the date in custom, odd format?
I've got a start script (bash) that should handle the output of the program, that contains dates with very odd formatting (like Jan, 4, 2021, 1:20:30 PM - which would be a pattern like "MMM, d, yyyy, h:mm:ss a".
It would be possible to extract the tokens with sed or awk, but processing them is a nightmare, especially month shortcuts (they are in the same language like system, but that language can be installation-specific) or hours (need to check AM/PM token and add 12 if it's PM).
'date' apparently doesn't support that, but maybe some shell extension or toolkit package? Or I'm out of luck and I need to parse this crappy format token by token with sed/cut/awk?
This what I've tried was to do touch -d "Date" after removing the commas (so that I can compare dates with [[ file1 -ot file2 ]], but the problem is, that touch has ignored the TIME part, and ls -lh has shown, that the year was set in place of time, and the result of the comparison was therefore invalid.
Convert date with GNU date and bash:
d="Jan, 4, 2021, 1:20:30 PM"
IFS=',:' read mon day year hour min sec a <<<"$d"
date -d "$mon $day $year $hour:$min:$sec $a" '+%Y-%m-%d %H:%M:%S'
Output:
2021-01-04 13:20:30
Related
I understand how to reformat a date using the date command, and I am fine with that. However, I have a wrinkle in that I am struggling with - the date I want to reformat is the output of another command, so I am storing it in the variable. I am struggling with the syntax of how to specify that I want to take the output of one command, and run it through date -d, and store it in another variable. Here is what I tried:
expdate=`get_expire_date.sh`
echo $expdate
Mon 23 Mar 2022 05:05:05 PM UTC
expdval=`date -d'($expdate)'`
echo $expdval
I get today's date, not the converted expire date from the script output. If I leave the parenthesis out, of course, it treats $expdate as the literal text to translate and gives an error, whereas if I leave the single quote marks off, it uses the spaces in the date string as a delimiter and only grabs the first token.
What am I doing wrong?
First, parameter expansion doesn't occur inside single quotes. You would need to change the single quotes
expdval=`date -d'($expdate)'`
to double quotes
expdval=`date -d"($expdate)"`
Second, the parentheses create an invalid input, which results (for reasons I don't really understand) in an output of midnight of the current day. (You'll get the same result with the trivial invalid date date -d "".)
Drop the parentheses, and you'll get the same date back (because the input format matches the default output format).
$ date -d "$expdate"
Wed Mar 23 13:05:05 EDT 2022
To actually manipulate it, you'll need an explicit output format:
$ date -d "$expdate" +%Y-%m-%d
2022-03-23
or some form of date "arithmetic":
$ date -d "$expdate + 2 days"
Fri Mar 25 13:05:05 EDT 2022
I found I had to use double-quotes instead, like this (and sorry for the old way of doing things, updating to new shell syntax):
expdval=$(date -d"$(get_expire_date.sh)")
date +'%A %B %d' | sed -e 's/\(^\|[^[:digit:]]\+\)0\+\([[:digit:]]\)/\1\2/g
I like the output of the above command, which strips leading zeroes off days of the month produced by the date command, in the case of numerals less than 10. It's the only way I've thus far found of producing single digit dates from the date command's output for the day of the month, which otherwise would be 01, 02, 03, etc.
A couple of questions in this regard. Is there a more elegant way of accomplishing the stated goal of stripping off zeroes? I do know about date's %e switch and would like to use it, but with numerals 10 and greater it has the undesirable effect of losing the space between the month name and the date (so, July 2 but July10).
The second question regards the larger intended goal of arriving at such an incantation. I'm putting together a script that will scrape some data from a web page. The best way of locating the target data on the page is by searching on the current date. But the site uses only single digits for the first 9 days of the month, thus the need to strip off leading zeroes. So what's the best way of getting this complex command into a variable so I can call it within my script? Would a variable within a variable be called for here?
RESOLUTION
I'll sort of answer my own question here, though it is really input from Renaud Pacalett (below) that enabled me to resolve the matter. His input revealed to me that I'd not understood very well the man page, particularly the part where is says "date pads numeric fields with zeroes," and below that where it is written "- (hyphen) do not pad the field." Had I understood better those statements, I would have realized that there is no need for the complex sed line through which I piped the date output in the title of this posting: had I used there %-d instead of just %d there would have been no leading zeroes in front of numerals less than 10 and so no need to call sed (or tr, as suggested below by LMC) to strip them off. In light of that, the answer to the second question about putting that incantation into a variable becomes elementary: var=$(date +'%A %B %-d') is all that is needed.
I may go ahead and mark Renaud Pacalet's response as the solution since, even though I did not implement all of his suggestions into the latest incarnation of my script, it proved crucial in clarifying key requirements of the task.
If your date utility supports it (the one from GNU coreutils does) you can use:
date +'%A %B %-d'
The - tells date to not pad the numeric field. Demo:
$ date -d"2021/07/01" +'%A %B %-d'
Thursday July 1
Not sure I understand your second question but if you want to pass this command to a shell script (I do not really understand why you would do that), you can use the eval shell command:
$ cat foo.sh
#!/usr/bin/env bash
foo="$(eval "$1")"
echo "$foo"
$ ./foo.sh 'date -d"2021/07/01" +"%A %B %-d"'
Thursday July 1
Please pay attention to the double (") and simple (') quotes usage. And of course, you will have to add to this example script what is needed to handle errors, avoid misuses...
Note that many string comparison utilities support one form or another of extended regular expressions. So getting rid of these leading zeros or spaces can be as easy as:
grep -E 'Thursday\s+July\s+0*1' foo.txt
This would match any line of foo.txt containing
Thursday<1 or more spaces>July<1 or more spaces><0 or more zeros>1
I'm having the hardest time trying to do the most simple thing. I want to print the current date with Year-month-day-hour-minute-seconds to a .csv file
I want the formatting of the .csv file to look like this:
Year Month Day Hour Minute Second
2017 03 08 17 52 23
And i cannot figure out for the life of me what to do. I have tried figuring it out with printf and using "," as the delimiter but the formatting will not work. It does not end up in the columns like i want.
Could someone point me in the right direction?
It's a simple date formatting problem. Try the following :
echo "Year,Month,Day,Hour,Minute,Second" && date +"%Y,%m,%d,%H,%M,%S" > myfile.csv
Here is a link to date formatting syntax.
I am running some gnu time scripts which generates output of the form
mm:ss.mm (minutes, seconds and miliseconds, for example 1:20.66)
or hh:MM:ss (hours, minutes and seconds, for example 1:43:38).
I want to convert this to seconds (in order to compare them and plot them in a graphic).
Which is the easiest way to do this using bash?
$ TZ=utc date -d '1970-01-01 1:43:38' +%s
6218
Assuming you can run the GNU date command:
date +'%s' -d "01:43:38.123"
If the script is generating "mm:ss.mm" you'll need to add "00:" to the beginning, or date will reject it.
If you're on a BSD system (including Mac OS X), you need to run date -j +'%s' "0143.38" unless you have GNU date installed with MacPorts or Homebrew or something.
And if you want pure Bash you can do something like
IFS=: read h m s <<<"${hms%.*}"
seconds=$((10#$s+10#$m*60+10#$h*3600))
The 10# part is mandatory to specify that the numbers are given in radix 10. Without this, you'd get errors if h, m or s is 08 or 09 (as Bash interprets numbers with a leading 0 in octal).
Is there a way to add date in the name of the file... we can add current date in this manner date '+%Y%m%d' but i want to add "filename_date_1-2-2011_thru_31-2-2011.txt" Is it possible to do that??????????
If you have a sufficiently advanced version of the date command and you know a Unix timestamp for the start and end dates, then you can use:
(MacOS X) date -r 1234567890 "+%d-%m-%Y" to obtain 13-02-2009.
(GNU) date -d 2/13/2009 "+%d-%m-%Y" to obtain 13-02-2009 again.
If you don't want the leading zeroes on the day of month, then you need to use '%e` instead of '%d' on Linux (but that puts a space in place of the zero). It is not clear that there's a format specifier for day-of-month without a leading zero on MacOS X; nor is it clear that there's a way to format month of year as a single-digit number for January to September on either platform.
You get the format into your C shell script using back-ticks around the date commands.
Consider reading Csh Programming Considered Harmful and heeding its advice.