Replace Newline With Space In Array Element In a For Loop In Bash

Question

I'm assigning values to an array inside a for loop:

aws_user_roles+=("$aws_role_name")

If I were assigning the values of the array from a command and want to strip the newline I could go:

 readarray -t aws_roles < <(...some aws commands...)

My for loop looks something like this:

for ((has_role_index=0;has_role_index<${#aws_user_has_roles[@]};++has_role_index)); do         
             aws_user_roles+=("$aws_role_name")
             declare -p aws_user_roles
done

How can I strip out the newline from the array elements in aws_user_roles and replace it with a space?

Do you really want to replace newlines with spaces, or do you want each line to be a separate array element? — Gordon Davisson
– Gordon Davisson, Commented Oct 30, 2018 at 18:35

KamilCuk · Accepted Answer · 2018-10-30 17:13:24Z

3

Use tr:

aws_user_roles+=("$(<<<"$aws_role_name" tr '\n' ' ')")

$(..) is command substitution
<<<"$variable" is a here-string
tr '\n' ' ' substitutes newlines for spaces.

answered Oct 30, 2018 at 17:13

KamilCuk

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Comments

oguz ismail · Accepted Answer · 2018-10-30 17:33:11Z

1

Use pattern substitution:

aws_user_roles+=("${aws_role_name//$'\n'/ }")

From man bash:

${parameter/pattern/string}

Pattern substitution. The pattern is expanded to produce a pattern just as in pathname expansion. Parameter is expanded and the longest match of pattern against its value is replaced with string. If pattern begins with /, all matches of pattern are replaced with string. Normally only the first match is replaced

answered Oct 30, 2018 at 17:33

oguz ismail

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