bash store output of command in array

Question

I'm trying to find if the output of the following command, stores just one file in the array array_a

array_a = $(find /path/dir1 -maxdepth 1 -name file_orders?.csv)
echo $array_a
/path/dir1/file_orders1.csv /path/dir1/file_orders2.csv
echo ${#array_a[@]}
1

So it tell's me there's just one element, but obviously there are 2. If I type echo ${array_a[0]} it doesn't return me anything. It's like, the variable array_a isn't an array at all. How can i force it to store the elements in array?

Add a shebang and then paste your script there: shellcheck.net — Cyrus
– Cyrus, Commented Jan 25, 2018 at 5:33

tripleee · Accepted Answer · 2018-01-25 05:35:27Z

3

You are lacking the parentheses which define an array. But the fundamental problem is that running find inside backticks will split on whitespace, so if any matching file could contain a space, it will produce more than one element in the resulting array.

With -maxdepth 1 anyway, just use the shell's globbing facilities instead; you don't need find at all.

 array_a=(/path/dir1/file_orders?.csv)

Also pay attention to quotes when using the array.

 echo "${array_a[@]}"

Without the quotes, the whitespace splitting will happen again.

answered Jan 25, 2018 at 5:35

tripleee

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