Arguments passed into for loop in bash script [duplicate]

Answer 1

As well as John Kugelman's solution, you can use eval like this:

x=10; for i in $(eval echo {1..$x}); do echo $i; done

Or, if $1 is 10, then:

set -- 10
for i in $(eval echo {1..$1})
do
    echo $i
done

You could also use some variants on:

set -- 1000
eval echo {1..$1} |
while read i
do
    echo $i
done

Or:

set -- 1000
while read i
do
     echo $i
done <(eval echo {1..$1})

That uses process substitution.

Answer 2

Variable substitutions are not done inside of curly braces. You can use fixed numbers but not variables.

Brace Expansion

A sequence expression takes the form {x..y}, where x and y are either integers or single characters. ...

Brace expansion is performed before any other expansions, and any characters special to other expansions are preserved in the result. It is strictly textual. Bash does not apply any syntactic interpretation to the context of the expansion or the text between the braces.

A correctly-formed brace expansion must contain unquoted opening and closing braces, and at least one unquoted comma or a valid sequence expression. Any incorrectly formed brace expansion is left unchanged.

Try one of these alternatives:

for ((i = 1; i <= $1; i++)); do
    echo $i
done

# Not recommended with large sequences.
for i in $(seq 1 $1); do
    echo $i
done

Answer 3

This will cycle through all true arguments (a.k.a. "testo mesto" is one argument)

#cycle through all args
for (( i=1; i<=$1; i++ )); do
    eval arg=\$$i
    echo "$arg"
done

OR

#cycle through all args
for (( i=1; i<=$1; i++ )); do
    echo "${!i}"
done

Answer 4

...or in the unlikely event that you really just want sequential numbers:

seq $1

:-)