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issues with bash commands as variables (monero.town)

submitted 1 year ago* (last edited 1 year ago) by shortwavesurfer@monero.town to c/programming

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Any ideas? I am attempting to write a script that uses sed.

If done this way it fails

rmdec="sed 's/..................$//'"
i1xmr=$(echo "$i1p/$apiresponse*1000" |bc -l |$rmdec)

But if i do it this way it works

i1xmr=$(echo "$i1p/$apiresponse*1000" |bc -l | sed 's/..................$//')

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[–] shortwavesurfer@monero.town 1 points 1 year ago (1 children)

Yeah, my goal is to shorten that sed command to that variable. It seems like it would work, but nope. It throws errors

[–] Andy@programming.dev 1 points 1 year ago (2 children)

It might be because it's a single string, and might work if you store it or expand it as an array. I think it would in Zsh, anyway.

But the response to use a function instead is probably wiser.

[–] brie 3 points 1 year ago

Strings work fine, the problem is the (single) quotes:

~ $ foo="echo 'hello world'"
~ $ for x in $foo; do echo $x; done
echo
'hello
world'
~ $ $foo
'hello world'
~ $ eval "$foo"
hello world

The splitting is by whitespace, so the single quotes remain in the arguments. Using eval (and double quotes to preven splitting), it gets processed correctly. That said, don't use eval; use functions or aliases instead.

[–] shortwavesurfer@monero.town 1 points 1 year ago

Yep, the function did the trick. My guess is it was being misread at execution as a variable and thats why it was breaking