Shell – Generate line number / “index” within a range from a date seed value

dategrepshell-script

I'm trying to create a bash script to display a word of the day (random for each day). I have a dictionary file that on each line has a word and its definition.

I'd like to use date to get a unique value for each day. Like so

today=$(date '+%Y%m%d') # will return 20160616 (for today)

Now I'd like to use this value to generate a line number for me to grab from the dictionary file.

My dictionary is 86036 lines long so I need to convert $today to a value between 1 and 86036.

What is the best way to do this?

Best Answer

A somewhat different solution: Scramble the lines in your text files with a cron job every day. Your script then picks the first line.

Cron job (requires a sort that can "sort" data randomly with -R):

0 0 * * * sort -R -o wotd_data.txt wotd_data.txt

Or, if your cron understands @daily (see man 5 crontab):

@daily sort -R -o wotd_data.txt wotd_data.txt

Script:

head -n 1 wotd_data.txt

Full paths to wotd_data.txt are obviously required.

Related Solutions

SED and GREP – Get All Regex Matches Between Two Patterns

IF GNU grep is an option, you could pass the -P (perl-compatible regex) flag and use lookahead assertions, lookbehind assertions and non-greedy matches to pull out what you need

echo 'xxSTART relevanttext xxEND something else xxSTART even more relevant'  |\
grep -oP '(?<=START).*?(?=xxEND|$)'
relevanttext
even more relevant

Or as Stephane Chazelas suggests, use the nifty \K in place of the look-behind assertion

echo 'xxSTART relevanttext xxEND something else xxSTART even more relevant'  |\
grep -oP 'START\K.*?(?=xxEND|$)'

Shell – How to parse ISO8601 dates with linux date command

You ask for "known workarounds." Here is a simple one:

$ date -d "$(echo 20140103T1422 | sed 's/T/ /')"
Fri Jan  3 14:22:00 PST 2014

This uses sed to replace "T" with a space. The result is a format that date understands.

If we add seconds onto the ISO8601 date, then date requires more changes:

$ date -d "$(echo 20140103T142211 | sed -r 's/(.*)T(..)(..)(..)/\1 \2:\3:\4/')"
Fri Jan  3 14:22:11 PST 2014

In the above, sed replaces the "T" with a space and also separates HHMMSS into HH:MM:SS.

Best Answer

Related Solutions

SED and GREP – Get All Regex Matches Between Two Patterns

Shell – How to parse ISO8601 dates with linux date command

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