oneliner 将输出附加到多个变量

Question

var=4
com="$(echo "A\nB\nC\nD" | sed 's/\\n/\n/g')"
seq "$var" | sed 's/^/var/g' | xargs  | sed 's/^/read /g' | sed "s/$/ <<< \"\$(echo \"\$com\")\"/g" 

Above outputs: read var1 var2 var3 var4 <<< "$(echo "$com")" What i want is to execute that command instead of outputing and echo all four set variables using && after it has been executed

Tried:

`seq "$var" | sed 's/^/var/g' | xargs  | sed 's/^/read /g' | sed "s/$/ <<< \"\$(echo \"\$com\")\"/g"` && echo $var1 $var2 $var3 $var4

Doesn't work

Tried:

seq "$var" | sed 's/^/var/g' | xargs  | sed 's/^/read /g' | sed "s/$/ <<< \"\$(echo \"\$com\")\"/g" | bash 

Above doesn't work because variables are set in seperate session not globally

Tried:

eval "seq "$var" | sed 's/^/var/g' | xargs  | sed 's/^/read /g' | sed \"s/$/ <<< \"\$(echo \"\$com\")\"/g"\"  && echo $var1 $var2 $var3 $var4

But doesn't work. Pleae help needed.

Desired output: After all variables set: ABCD

seq "$var" | sed 's/^/var/g' | xargs  | sed 's/^/read /g' | sed "s/$/ <<< \"\$(echo \"\$com\")\" \&\& echo \$var1 \$var2 \$var3 \var4/g" | bash

Above may work but $com is not expanding inside sed

Answer 1

Finally Got It Worked

var=4
com="$(echo "A\nB\nC\nD" | sed 's/\\n/\n/g')"
seq "$var" | sed 's/^/var/g' | xargs  | sed 's/^/read /g' | sed "s/$/ <<< \"\$(echo `echo $com`)\" \&\& echo \$var1 \$var2 \$var3 \$var4/g" | bash

Answer 2

Finally Solved

var=4
com="$(echo "A\nB\nC\nD" | sed 's/\\n/\n/g')"
eval "seq "$var" | sed 's/^/var/g' | xargs  | sed 's/^/read /g' | sed \"s/$/ <<< \$\(echo `echo $com`\) \&\& echo \\\$var1 \\\$var2 \\\$var3 \\\$var4/g\"" | bash

Answer 3

Assumptions:

• objective is to define and populate four new variables: var1=A , var2=B , var3=C and var4=D
• optionally export the variables

Generating the variable names and values:

$ $ printf "var%s\n" {1..4}
var1
var2
var3
var4

$ printf "%s\n" {A..D}
A
B
C
D

Associate variable names with values:

$ paste <(printf "var%s\n" {1..4}) <(printf "%s\n" {A..D})
var1    A
var2    B
var3    C
var4    D

Short demo of a bash nameref :

$ typeset -p myvar
-bash: typeset: myvar: not found

$ declare -n newvar="myvar"
$ newvar='new_value'

$ typeset -p newvar myvar
declare -n newvar="myvar"
declare -- myvar="new_value"

$ echo "${newvar}"
new_value

$ echo "${myvar}"
new_value

From here we can use a while loop and a nameref to create and populate the new variables:

# verify variables do not exist:

$ typeset -p var1 var2 var3 var4
-bash: typeset: var1: not found
-bash: typeset: var2: not found
-bash: typeset: var3: not found
-bash: typeset: var4: not found

# create/populate/export new variables

while read -r name val
do
    declare -n x="${name}"
    export x="${val}"
done < <(paste <(printf "var%s\n" {1..4}) <(printf "%s\n" {A..D}))

# verify variables now exist, are populated and exported (-x):

$ typeset -p var1 var2 var3 var4
declare -x var1="A"
declare -x var2="B"
declare -x var3="C"
declare -x var4="D"