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    <p><strong>For GNU <code>date</code>:</strong></p> <pre><code>date_222days_before_TodayYear=$(date --date="222 days ago" +"%Y") date_222days_before_TodayMonth=$(date --date="222 days ago" +"%m") date_222days_before_TodayDay=$(date --date="222 days ago" +"%d") </code></pre> <hr> <p><strong>For BSD <code>date</code>:</strong>: </p> <p>If you are using OS X or FreeBSD, use the following instead because BSD date is different from GNU date:</p> <pre><code>date_222days_before_TodayYear=$(date -j -v-222d +"%Y") date_222days_before_TodayMonth=$(date -j -v-222d +"%m") date_222days_before_TodayDay=$(date -j -v-222d +"%d") </code></pre> <p>Source: <a href="http://www.unix.com/man-page/FreeBSD/1/date/" rel="noreferrer">BSD date manual page</a></p> <p><strong>Note:</strong></p> <p>In <code>bash</code> and many other languages, you cannot start a variable name with a numerical character, so I prefixed them with <code>date_</code> for you.</p> <hr> <p><strong>Second Update:</strong> New requirement - <em>Using 222 Working days instead of 222 Regular days:</em></p> <p>(<strong>Assumption:</strong> Not considering statutory holidays, because that just gets far beyond the scope of what I can help you with in a shell script:)</p> <p>Consider 222 working days:</p> <ul> <li>5 working days per week, that is <code>floor(222/5) == 44 weeks</code></li> <li><code>44 weeks * 7 days per week == 308 days</code></li> <li>Extra days leftover: <code>222 % 5 == 2</code></li> <li>Therefore <code>222 working days == 310 regular days</code></li> </ul> <p>But, there is a catch! If the number of regular days is <code>308</code> or some multiple of <code>7</code>, then we would have been fine, because any multiple of 7-days ago from a working day is still a working day. So we need to consider whether today is a Monday or a Tuesday:</p> <ul> <li>If today is a Monday, we'd get Saturday where we wanted Thursday</li> <li>If today is a Tuesday, we'd get Sunday where we wanted Friday</li> </ul> <p>So you see we need an additional offset of 2 more days if today is either Monday or Tuesday; so let's find that out first before we proceed:</p> <pre><code>#!/bin/bash # Use 310 days as offset instead of 222 offset=310 # Find locale's abbreviated weekday name (e.g., Sun) today=$(date -j +"%a") # Check for Mon/Tue if [[ "$today" == "Mon" ]] || [[ "$today" == "Tue" ]]; then offset=$((offset+2)) fi date_222_working_days_before_TodayYear=$(date -j -v-${offset}d +"%Y") date_222_working_days_before_TodayMonth=$(date -j -v-${offset}d +"%m") date_222_working_days_before_TodayDay=$(date -j -v-${offset}d +"%d") </code></pre> <p>And that should do it =)</p>
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    1. COI get `date: illegal date format` - I wonder if I might be using the wrong type of quotation marks (I copied and pasted from here but that script errored). Any ideas?
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    2. CO@user1644609 try the updated answer, but the quotes before shouldn't be the error. Your question is tagged `bash`, but do you know which version of shell you are using?
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    3. COThanks @sampson-chen - I've tried your update, but now, instead of getting `illegal date format` I get `illegal time format` (and then usage instructions). My bash version is: `GNU bash, version 3.2.48(1)-release (x86_64-apple-darwin12) Copyright (C) 2007 Free Software Foundation, Inc.`
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