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ajeetdsouza avatar ajeetdsouza commented on June 8, 2024 1

My bad, I missed the $ sign. This function will work for you:

function z() {
  z "$@"
  ls
}

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ajeetdsouza avatar ajeetdsouza commented on June 8, 2024

Have you tried z "@"?

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nikitarevenco avatar nikitarevenco commented on June 8, 2024

Have you tried z "@"?

It says there is no match found. However, I was able to make it simpler with:

function s() { 
  z $1
  ls
}

(still doesn't change the original idea though)

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