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Bash function to see if the day ends in "y"

Terminal - Bash function to see if the day ends in "y"
function ends_in_y() { if [ `date +%A | sed -e 's/\(^.*\)\(.$\)/\2/'` == "y" ]; then echo 1; else echo 0; fi }
2010-04-06 20:14:34
User: allrightname
Functions: echo sed
-3
Bash function to see if the day ends in "y"

For those days when you need to know if something is happening because the day ends in "y".

Alternatives

There is 1 alternative - vote for the best!

Terminal - Alternatives
date +%A | cut -c $(( $(date +%A | wc -c) - 1 ))
2010-04-07 00:23:15
User: DaveQB
Functions: cut date wc
Tags: bash echo cut date wc
0

A command to find out what the day ends in. Can be edited slightly to find out what "any" output ends in.

NB: I haven't tested with weird and wonderful output.

function ends_in_y() { case $(date +%A) in *y ) true ;; * ) false ;; esac } ; ends_in_y && echo ok
2010-04-06 22:18:52
Functions: date echo false true
-1

The shell has perfectly adequate pattern matching for simple expressions.

Know a better way?

If you can do better, submit your command here.

What others think

Silly as it is, there are shorter ways to get there.

You don't need the "-e", or the first () in sed.

sed 's/^.*\(.$\)/\1/'

But I prefer grep:

function ends_in_y() { date +%A | grep -qv "y$"; echo $?; }
Comment by flatcap 281 weeks and 2 days ago
true
Comment by penpen 281 weeks and 2 days ago

And now for the ultimate solution:

:
Comment by penpen 281 weeks and 2 days ago

for future reference, if you're grabbing some regular expression out of a line of text and discarding the rest (in this case, ".$") grep -o works wonders

Comment by camocrazed 267 weeks and 2 days ago

Your point of view

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