how can i do echo this way ?

[AntonioPt]

hello all i need some help
how do i achive this ? thxx in advance

Code: Select all

# let make 2 var
var_1="this is var 1"
var_2="this is var 2"
# now i wanna call them in loop
for i in {1..2} ; do
  echo "var_$i"
done

# i want this output
this is var 1
this is var 2

[rcrsn51]

Code: Select all

eval echo \$var_$i

[AntonioPt]

Code: Select all

eval echo \$var_$i

thxxxxxxxxxxx

[mow9902]

hello all i need some help
how do i achive this ? thxx in advance

Code: Select all

# let make 2 var
var_1="this is var 1"
var_2="this is var 2"
# now i wanna call them in loop
for i in {1..2} ; do
  echo "var_$i"
done

# i want this output
this is var 1
this is var 2

change echo "var_$i" to
echo "This is var_$i"

[AntonioPt]

hello all i need some help
how do i achive this ? thxx in advance

Code: Select all

# let make 2 var
var_1="this is var 1"
var_2="this is var 2"
# now i wanna call them in loop
for i in {1..2} ; do
  echo "var_$i"
done

# i want this output
this is var 1
this is var 2

change echo "var_$i" to
echo "This is var_$i"

Thank you but that wasn't what i wanna

[MochiMoppel]

Another way:

Code: Select all

var_1="this is var 1"
var_2="this is var 2"
for i in {1..2} ; do
  i=var_$i
  echo "${!i}"
done

[AntonioPt]

Another way:

Code: Select all

var_1="this is var 1"
var_2="this is var 2"
for i in {1..2} ; do
  i=var_$i
  echo "${!i}"
done

I like that thxx

[MochiMoppel]

Code: Select all

eval echo \$var_$i

thxxxxxxxxxxx

This only appears to work because your example values contain only single spaces.
Will not work with other or with continuous whitespace

Code: Select all

var_1='this   is   var 1'
var_2='this		is     var 2'
for i in {1..2} ; do
  eval echo \$var_$i
done

[AntonioPt]

Code: Select all

eval echo \$var_$i

thxxxxxxxxxxx

This only appears to work because your example values contain only single spaces.
Will not work with other or with continuous whitespace

Code: Select all

var_1='this   is   var 1'
var_2='this		is     var 2'
for i in {1..2} ; do
  eval echo \$var_$i
done

ahhhhh thankyou for let me know about this small detail very important indeed just update with your code more clean

[rcrsn51]

This only appears to work because your example values contain only single spaces.
Will not work with other or with continuous whitespace

Code: Select all

eval echo \"\$var_$i\"

[superhik]

Instead of eval, printf, IFS and cat are options that work.

Code: Select all

var_1='this   is   var 1'
var_2='this		is     var 2'  # Contains tabs and multiple spaces

for i in {1..2}; do
  var_name="var_$i"
  printf "%s\n" "${!var_name}"  # Preserves all spaces and tabs
done

for i in {1..2}; do
  var_name="var_$i"
  IFS= read -r value <<< "${!var_name}"  # Read the variable while preserving spaces
  echo "$value"
done

for i in {1..2}; do
  var_name="var_$i"
  cat <<< "${!var_name}"
done

Also echo -E works in bash

Code: Select all

for i in {1..2}; do
  var_name="var_$i"
  echo -E "${!var_name}"
done

Best universal options: printf "%s\n" or IFS= read -r

[MochiMoppel]

Also echo -E works in bash

This was already one of the proposed solutions - withouth the explicit -E option. As this is the default in echo anyway and therefore not required, do you see any scenario where the -E option makes a difference?

Best universal options: printf "%s\n" or IFS= read -r

What makes them "best"? They also work only with bash.

[superhik]

Actually, IFS and printf are not Bash-specific, they both work in POSIX compliant shells.
Unlike echo, which may collapse spaces and interpret special characters differently across systems, printf always prints exactly what you pass to it.
printf "%s\n" always preserves spaces/tabs → Works in sh, bash, dash, ksh, etc.
The default read behavior trims spaces, but setting IFS='' prevents this.

Code: Select all

var='   hello   world   '
IFS= read -r value <<EOF
$var
EOF
printf "%s\n" "$value"  # Preserves spaces exactly

IFS works in all POSIX shells, including sh, dash, ksh, etc.
cat <<< is bash only.
eval "echo \${var}" Might collapse, and if var contains malicious code (e.g., var='$(rm -rf /)'), eval will execute it!
eval "echo \"\${var}\"" preserves spaces, but still risky.
printf "%s\n" "$var" preserves spaces and it's safe. Therefore it's the best universal solution.

Also echo -E works in bash

This was already one of the proposed solutions - withouth the explicit -E option. As this is the default in echo anyway and therefore not required, do you see any scenario where the -E option makes a difference?

You're absolutely right! In modern Bash, echo -E is redundant because -E is the default behavior.
However, there are a few scenarios where echo -E can make a difference:
In POSIX sh, echo behavior varies by implementation. Some shells (like dash or BusyBox sh) might still interpret escape sequences by default, so using -E explicitly in Bash ensures consistency when running scripts across different shells.
In Bash, enabling shopt -s xpg_echo makes echo behave like echo -e, where \n and other escapes are interpreted by default.
Using echo -E here prevents this behavior.
If a user has an alias or function overriding echo in their shell profile (~/.bashrc, ~/.profile), echo might behave differently.
alias echo='echo -e'
Now, echo "Hello\nWorld" would always interpret \n as a newline, even in Bash.
Using echo -E forces literal output, ignoring the alias.

[MochiMoppel]

Actually, IFS and printf are not Bash-specific, they both work in POSIX compliant shells.

Sure. I was rather referring to the whole constructs in which they are used. These constructs are Bash-specific, using "indirect expansion" (${!var_name}). Which may, after some changes to the loop, leave the eval evil as the only POSIX compliant solution