Just annotations of little "how to's", so I know I can find how to do something I've already done when I need to do it again, in case I don't remember anymore, which is not unlikely. Hopefully they can be useful to others, but I can't guarantee that it will work, or that it won't even make things worse.
Bash quickie] Get a "random" number (larger than zero, within a range) with "shuf" only
Posted 07-31-2014 at 01:16 AM by the dsc
Updated 07-31-2014 at 01:39 AM by the dsc (deweirdifying the wording)
Updated 07-31-2014 at 01:39 AM by the dsc (deweirdifying the wording)
It always annoys me when I see things like "cat file | grep pattern | sed ..." when just "sed" would suffice, but I was guilty of using seq only to shuf random numbers when shuf can do it by itself.
shuf -n 1 -i 0-999999999
(apparently the largest number you can get is $((9999**8)) (9999^8 in bash-math), larger than that it gets negative, for some reason )
You also don't need to do some fancy "echo -e" with several "\n" (new lines) to shuf words, or anything else you could have imagined (like getting the number of variables in the array and "formatting" the echo to shuf a given number -- guilty too). Instead one can do:
shuf -n 1 -e foo bar whatever who-knew I-did-not yaddayadda "this is a phrase but counts as a word within this scheme" etc etc
shuf -n 1 -i 0-999999999
(apparently the largest number you can get is $((9999**8)) (9999^8 in bash-math), larger than that it gets negative, for some reason )
You also don't need to do some fancy "echo -e" with several "\n" (new lines) to shuf words, or anything else you could have imagined (like getting the number of variables in the array and "formatting" the echo to shuf a given number -- guilty too). Instead one can do:
shuf -n 1 -e foo bar whatever who-knew I-did-not yaddayadda "this is a phrase but counts as a word within this scheme" etc etc
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