Script for counting weekends - shell

Script for counting weekends (saturdays AND sundays) in a given month, by passing in month and year. How to write it?
New in scripting, can't find solution to this easy task.

You want to count weekends Sat and Sun, So I am guessing you mean 2 day weekends.
Hope this helps. VonBell
In unix/linux you can easily parse the cal command for the given month.
$ cal 04 2013
April 2013
Su Mo Tu We Th Fr Sa
... 1 2 3 4 5 6
7 8 9 10 11 12 13
14 15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30
The numerical date always start on Line 3, the cols for each day are constant
When I used the commands below to extract the weekend (sun, Sat)
As you can see Sun is cols 1,2,3 (with space included) and Sat is cols 19,20
$ cal 04 2013|tail +3|cut -c1-3,19,20
.. 6
7 13
14 20
21 27
28
When I execute the script below as cntWkEnd.sh The output is as follows
1 equals one day wkend and 2 equals 2 day wkend sat and sun
$ ./cntWkEnd.sh 04 2013
1
2
2
2
1
0
To count only 2 day weekends you can add at the command line or in the script
the following
./cntWkEnd.sh 04 2013|grep "2"|wc -l (Output of 3 shown below)
3
This the contents of cntWkEnd.sh
#!/bin/bash
cal $1 $2|tail +3|cut -c1-3,19,20 |\
while read WkEnd
do
echo $WkEnd|wc -w
done

Related

Why doesn't the output of the command `cal` have the day highlighted?

I need that the output of the command cal stay highlighted in a variable in order print it into a wallpaper so I can disable the calendar software to save resources of a low spec computer. How can I make the cal output stay highlighted?
Perhaps something like:
cal -h | sed 's/ \('"$(date +%_d)"'\) /{\1}/'
outputs
June 2021
Su Mo Tu We Th Fr Sa
{ 1} 2 3 4 5
6 7 8 9 10 11 12
13 14 15 16 17 18 19
20 21 22 23 24 25 26
27 28 29 30

How to convert week number and week day into date

I have CVS files which contain date information in three separate colums which I would like to combine. The information I have is:
Two digit year (field 2)
Week number (field 3)
Day of week number (field 4)
How can I convert these 3 numbers into normal date format of the form YYYYMMDD?
My input file looks like:
740054,17,40,1,0000000000001,25,25,test1,1
740054,17,40,2,0000000000001,24,24,test2,1
740054,17,40,4,0000000000001,19,19,test3,1
And the expected output I would like to have is:
740054,20171002,0000000000001,25,25,test1,1
740054,20171003,0000000000001,24,24,test2,1
740054,20171005,0000000000001,19,19,test3,1
As an example for the first line: October 2, 2017 is the Monday (1) of the 40th week of the year 2017
Does anybody know how to do such a conversion?
I will make the assumption that your week-number is according to the ISO 8601 definition (for other definitions see here). This ISO 8601 standard is widely used in the world: EU and most of other
European countries, most of Asia, and Oceania
The ISO 8601 standard states the following:
There are 7 days in a week
The first day of the week is a Monday
The first week is the first week of the year which contains a
Thursday. This means it is the first week with 4 days or more
in January.
With this definition, it is possible to have a week number 53. These occur with the first of January is on a
Friday (E.g. 2016-01-01, 2010-01-01). Or, if the year before was a
leap year, also a Saturday. (E.g. 2005-01-01)
December 2015 January 2016
Mo Tu We Th Fr Sa Su CW Mo Tu We Th Fr Sa Su CW
1 2 3 4 5 6 49 1 2 3 53
7 8 9 10 11 12 13 50 4 5 6 7 8 9 10 01
14 15 16 17 18 19 20 51 11 12 13 14 15 16 17 02
21 22 23 24 25 26 27 52 18 19 20 21 22 23 24 03
28 29 30 31 53 25 26 27 28 29 30 31 04
Given the year, week_number and day_of_week, how can we reconstruct the date? The answer requires several steps and will compute the day of the year (doy) of the requested date.
To compute the day of the year doy we first need to figure out when the first-week starts as explained above. If Jan 01 is a Tuesday, then the first week only contains 6 days and not 7, while if Jan 01 is a Friday, the first week starts only the week after. So we can solve this by adding an offset. The offset can be found in the following table:
dow001 str: Mo Tu We Th Fr Sa Su
dow001 num: 01 02 03 04 05 06 07
offset : 0 -1 -2 -3 3 2 1
and this offset is computed as 3-(dow001+2)%7
So with this, the day of the year is very easily computed:
doy = (week_number-1) * 7 + 3-(dow001+2)%7 + day_of_week
So having this, we can write the following GNU awk tool:
awk 'function compute_date(YYYY,CW,DOW) {
dow001 = strftime("%u",mktime(YYYY " 01 01 00 00 00"))
doy = (CW-1)*7 + (3 - (dow001+2)%7) + DOW
return strftime("%Y%m%d",mktime(YYYY " 01 " doy " 00 00 00"))}
}
BEGIN { FS = OFS = "," }
{ datestr = compute_date(2000+$2,$3,$4) }
{ print $1, datestr , $5,$6,$7,$8,$9 }' file
740054,20171002,0000000000001,25,25,test1,1
740054,20171003,0000000000001,24,24,test2,1
740054,20171005,0000000000001,19,19,test3,1

Shell Script to print a calendar after the user specifies the month and a day

*****Shell Script*******
Given a month and the day of the week that's the first of that month, print a calendar for the month. (Remember, number of days in months is different and use \n to go to a new line.)
Unix has a cal command especially for this purpose.
By default, cal shows the current month's calendar.
mayankp#mayank:~/$ cal
November 2018
Su Mo Tu We Th Fr Sa
1 2 3
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30
If you want a calendar for a specific month of a specific year, do this:
mayankp#mayank:~/$ cal 1 2018
January 2018
Su Mo Tu We Th Fr Sa
1 2 3 4 5 6
7 8 9 10 11 12 13
14 15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30 31
This displays the calendar for January 2018.
So, your shell script would be:(ex: calendar.sh)
#!/usr/bin/env bash
month=$1
year=$2
cal $1 $2
Run the script like this:
mayankp#mayank:~/$ sh calendar.sh 3 2018
March 2018
Su Mo Tu We Th Fr Sa
1 2 3
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30 31
Let me know if this helps.

List all the mondays of this month

I'm pretty new to bash and all the terminal in general - I've been messing around with cal and the date scripts wondering if there is anyway to list all the dates of monday of the current month .
My thought process is going thru the cal command, listing out the dates and maybe cutting a column from that input. Is that possible ?
You can do it with date command. Print 10 mondays since month ago:
for x in $(seq 0 9)
do
date -d "$x monday 5 week ago"
done
And grep only current month. Full command: for x in $(seq 0 9); do; date -d "$x monday 5 week ago"; done | grep $(date +%b)
Output:
Mon Jun 5 00:00:00 MSK 2017
Mon Jun 12 00:00:00 MSK 2017
Mon Jun 19 00:00:00 MSK 2017
Mon Jun 26 00:00:00 MSK 2017
Given:
$ cal
June 2017
Su Mo Tu We Th Fr Sa
1 2 3
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30
You can do:
$ cal | awk 'NF>5{print $2}'
Mo
5
12
19
26
If you want something that will support any day of cal, use the field width (gawk only this):
$ cal | gawk -v n=5 '
BEGIN{
FIELDWIDTHS = "3 3 3 3 3 3 3"
}
FNR>1{print $n}'
Th
1
8
15
22
29
Or, as pointed out in comments:
$ ncal | awk '/^Mo/'
Mo 5 12 19 26
Combination of cal,cut commands to achieve the output.
cal -h| cut -c'4,5'
Remove the highlight and cut the characters which suits in the fields of monday.
ncal | sed -n '/^Mo/p'
The output as below:
Mo 5 12 19 26

unix cal command special character

When I try "cal | tail -6" in my unix machine, I get -
1 2 3
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30
but when I try "cal | tail -6 | awk '{print $7}'", I get -
10
17
24
where is 3 going ? My requirement is basically all weekdays i.e column 2,3,4,5 & 6.
But I'm getting wrong output because of the strange behavior of "cal"
There are only 3 whitespace delimited columns in your first row. cal is working exactly as corrected, you are not understanding how awk works. As far as awk is concerned there is no 7th column in your first row as it yields attention to whitespace delimited columns, not fixed width columns.
A quick google search reveals you can use
BEGIN { FIELDWIDTHS = "3 3 3 3 3 3 3" }
In your awk script.
Since all of your columns in each row are three characters wide, you could use this to extract the days you wish for. For example, if you wanted only the 7th day in a column, you could do the following:
cal | sed 's/^\(.\{18\}\).*$/\1/'
This command would remove the first 18 characters in the line, which are the entries for the first 6 days of the week.
To extract a particular day, such as the fourth day, you could do this:
cal | sed 's/^.\{9\}\(.\{3\}\).*$/\1/'
To remove the first day of the week and the last day, you could do this:
cal | sed -e 's/^.\{3\}//' -e 's/^\(.\{15\}\).\{3\}$/\1/'
May be a row-wise extraction will do the trick. Try ncal. For example:
$ ncal
November 2012
Mo 5 12 19 26
Tu 6 13 20 27
We 7 14 21 28
Th 1 8 15 22 29
Fr 2 9 16 23 30
Sa 3 10 17 24
Su 4 11 18 25
or fill the absent dates with place holder (with '-' for example):
kent$ cal -s|tail -6|awk 'NR==1&&NF<7{gsub(/^ */,"");for(i=1;i<=(7-NF);i++) x=" - "x;$0=x" "$0;}1'
- - - - 1 2 3
4 5 6 7 8 9 10
11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30
then you could get the column, replace '-' with " " if needed. e.g. for $7:
kent$ cal -s|tail -6|awk 'NR==1&&NF<7{gsub(/^ */,"");for(i=1;i<=(7-NF);i++) x=" - "x;$0=x" "$0;}{print $7}'
3
10
17
24
Note that todays date is highlighted unless you turn it off (-h). Use cut to extract the wanted columns:
cal -h | cut -c19-20
Output:
Sa
3
10
17
24

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