Get the previous month from date

Given two numbers, month and year, how can I compute the first and the last day of that month ? My goal is to output these three lines:

1/ month / year (month in textual form but that is trivial)

2/ for each day of the month: name of the day of the week for the current day: Fri. & Sat. & Sun. [...]

3/ day number within the month: 1 & 2 & 3 [...] & 28 & ..

We have year folder say in a path /opt/informat/Archive a folder 2012. And in the same folder /opt/informat/Archive we have different folders month based, like 201210, 201211, 201212, this have data for each month, ie files. Now time to time i need to move the monthly folders to the main folder 2012 and zip it. Can please some one assist me with a script for this?

Hi, I need subtract two date values (which are in day of the year format) and the output would give the remaining days. using the command date +"%j" i would get today's 'day of the year' i.e.,

Code: > date +"%j" 256 Next, i need to take input of a previous date in the format 09/05/2012 and then convert it to 'day of the year format'.

I need a script to search for *.LOG in a directory, output the line after the current date and yesterday's date and also output the next 5 lines after the current date and yesterday's date

I have tried following commands, still no result

Code: Currentdate=`date -u +"%m/%d/%y"` awk '/$Currentdate/{getline; print}' *.LOG The log has data for the whole month for which I need only the

In a database I have saved the date in form of 3 columns: Year, Month, Day_of_Month ( I know it might not be a best way to save the date but I make certain other queries for which this format felt suitable.)

Now I wish to get all the rows where the date is in between two specified dates.

Dear experts,

I'm using solaris 5.10 and bash. I want to zip file "Amount.txt" to "Amount.zip" and rename it to "Amount_<prev_month>_<this year>.zip". For example, file for this month should be renamed to "Amount_06_2012.zip", for next month it should be "Amount_07_2012.zip". I have no problem for zipping the file.

hi guys,

i m new to shell script..

i dont know how to type cast string into integers and i mention my problem below..

date="21091988" month=$((${date:2:2})) # extract the month from previous date variable temp=$(($month*23))

when i am trying to calculate the temp value i got the following error.. 09: value too great for base (error token is "09") *23: syntax error:

How to convert date format such as 7/18/2015 to the number of month from requesting date 'date' in sh scripting ?

Let say I have output in my log.txt -> 7/18/2015. How I convert it to the full number of month starting from 'date' till 7/18/2015 in shell scripting ? Thanks in advance.

I am trying to get Month and Date from Date in Linux. this is my code

# Set Date D="2013/01/17"

# get day DD=$(D+"%d")

# get day MM=$(D+"%M")

# Day echo "Day:"$DD echo "Month:"$MM