shell script assigning value to a variable that contains a variable

Question

I'm planning to create a for loop for this one. But before that, I'm trying to simulate if I can assign a value in a variable that has a variable in it. See below:

x="5"
$myvar${x}=$(ls -lrt FILE_NAME_* | sed -n ${x}p | cut -c47-59)

echo $myvar5

I wanted to assign the filename value in $myvar5 but I don't think I'm using my variables correctly.

[Why you shouldn't parse the output of ls(1)](https://mywiki.wooledge.org/ParsingLs) — ceving, Oct 20 '21 at 07:54
@SomeTingWong: Since you are using _bash_, `myvar` could be an array. An alternative would be to use a _nameref_ (see the bash man-page), depending on what you want to do in the end. — user1934428, Oct 20 '21 at 08:05

score 1 · Accepted Answer · answered Oct 20 '21 at 08:02

1

What you are talking about is called indirection. But this works only for right hand values.

For left hand values you can use declare:

declare myvar$x=foo

answered Oct 20 '21 at 08:02

ceving

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Thank you! This does what I'm trying to accomplish :D – Some Ting Wong Oct 20 '21 at 08:19

shell script assigning value to a variable that contains a variable

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