How to use variables in a bash for loop [duplicate]

Answer 1

Brace subtitutions are performed before any other, so you need to use eval or a third-party tool like seq.

Example for eval:

for i in `eval echo {0..$length}`; do echo $i; done

This information can actually be found in man bash:

A sequence expression takes the form {x..y[..incr]}, where x and y are either integers or single characters, and incr, an optional increment, is an integer. [...]

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.

Answer 2

One way is using eval:

for i in $( eval echo {0..$length} )
do
       echo "do something right $i"
done

Note what happens when you set length=;ls or length=; rm * (don't try the latter though).

safely, using seq:

for i in $( seq 0 $length )
do
       echo "do something right $i"
done

or you can use the c-style for loop, which is also safe:

for (( i = 0; i <= $length; i++ )) 
do 
       echo "do something right $i"
done

Answer 3

In bash, brace expansion is the first step attempted so, at that point, $length will not have been substituted.

The manpage for bash states clearly:

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

There are a number of possibilities, such as using:

pax> for i in $(seq 0 $length) ; do echo $i ; done
0
1
2
3

though that may give you a large command line if length is massive.

Another alternative is to use the C-like syntax:

pax> for (( i = 0; i <= $length; i++ )) ; do echo $i; done
0
1
2
3