bash: pass multiple parameters from a single variable

Question

In a bash call I want to put some constant parameters to a variable and don't lose StdOut & StdErr inside a pipe.

I have a call

git fetch origin "ref1:ref1" "ref2:ref2" "ref3:ref3"

Let's I'll put those constant values to a variable

fetch_refspec="'ref1:ref1' 'ref2:ref2' 'ref3:ref3'"

I see a solution to use a pipe, but I'm afraid to lose the output somehow. And I do not want to use files for a caching (tee command).

echo $refs | xargs git origin

I don't understand how to do this stuff cleverly. Or if it possible at all.
Later I want to put the output to a variable and analyze it.

Use array as fetch_refspec="(ref1:ref1' 'ref2:ref2' 'ref3:ref3'); git fetch origin "${fetch_refspec[@]}" — anubhava
– anubhava, Commented Sep 18, 2017 at 11:31

Noelle L. · Accepted Answer · 2024-12-30 12:07:15Z

9

Don't use a variable, use an array!

declare -a gitArgs=("ref1:ref1" "ref2:ref2" "ref3:ref3")

and pass it to the command you need,

git origin "${gitArgs[@]}"

edited Dec 30, 2024 at 12:07

Noelle L.

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answered Sep 18, 2017 at 11:32

Inian

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2 Comments

it3xl Over a year ago

Got harsh bugs :) with Array variables may not (yet) be exported. But OK with it. Forced to use source command.

it3xl Over a year ago

On the other hand, using of the source command to call subscripts kills preconditions. I.e. the shortest form if [[ ... ]]; then exit at beginning of subscripts now must be converted and wraped over entire script. And I have many preconditions ))

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