I have gotten the following to work:
for i in {2..10}
do
echo "output: $i"
done
It produces a bunch of lines of output: 2, output: 3, so on.
However, trying to run the following:
max=10
for i in {2..$max}
do
echo "$i"
done
produces the following:
output: {2..10}
How can I get the compiler to realize it should treat $max as the other end of the array, and not part of a string?
Brace expansion, {x..y} is performed before other expansions, so you cannot use that for variable length sequences.
Instead, use the seq 2 $max method as user mob stated.
So, for your example it would be:
max=10
for i in `seq 2 $max`
do
echo "$i"
done
seq to count and increment numbers in the for loop, hence it is recommend that you avoid using seq. This command is left for compatibility with old bash. The built-in commands are fast enough. for (( EXP1; EXP2; EXP3 )) ...
for (( ... )) syntax isn't POSIX and that means for example that it won't work out of the box on things like Alpine Linux. seq does.
#!/bin/sh is Dash in Debian/Ubuntu.Try the arithmetic-expression version of for:
max=10
for (( i=2; i <= $max; ++i ))
do
echo "$i"
done
This is available in most versions of bash, and should be Bourne shell (sh) compatible also.
sh, instead making it a link to other another shell. Ideally, such a shell invoked as sh would only support those features in the POSIX standard, but by default let some of their extra features through. The C-style for-loop is not a POSIX feature, but may be in sh mode by the actual shell.Step the loop manually:
i=0
max=10
while [ $i -lt $max ]
do
echo "output: $i"
true $(( i++ ))
done
If you don’t have to be totally POSIX, you can use the arithmetic for loop:
max=10 for (( i=0; i < max; i++ )); do echo "output: $i"; done
Or use jot(1) on BSD systems:
for i in $( jot 0 10 ); do echo "output: $i"; done
true $(( i++ )) doesn't work in all cases, so most portable would be true $((i=i+1)).
jot 0 10 might be an infinite loop starting at 10. To get the same effect as max=10; for (( i=0; i < max; i++ )); do echo "output: $i"; done I think the syntax should be jot -w 'output: %d' 10 0 (directly print out sans loop) or just for i in $( jot 10 0 ); do ….. for the looped variantIf the seq command available on your system:
for i in `seq 2 $max`
do
echo "output: $i"
done
If not, then use poor man's seq with perl:
seq=`perl -e "\$,=' ';print 2..$max"`
for i in $seq
do
echo "output: $i"
done
Watch those quote marks.
We can iterate loop like as C programming.
#!/bin/bash
for ((i=1; i<=20; i=i+1))
do
echo $i
done
There's more than one way to do it.
max=10
for i in `eval "echo {2..$max}"`
do
echo "$i"
done
eval will evaluate anything you set max to. Consider max="2}; echo ha; #", then replace echo ha with something more destructive.
This is a way:
Bash:
max=10
for i in $(bash -c "echo {2..${max}}"); do echo $i; done
The above Bash way will work for ksh and zsh too, when bash -c is replaced with ksh -c or zsh -c respectively.
Note: for i in {2..${max}}; do echo $i; done works in zsh and ksh.
Well, as I didn't have the seq command installed on my system (Mac OS X v10.6.1 (Snow Leopard)), I ended up using a while loop instead:
max=5
i=1
while [ $max -gt $i ]
do
(stuff)
done
*Shrugs* Whatever works.
These all do {1..8} and should all be POSIX. They also will not break if you
put a conditional continue in the loop. The canonical way:
f=
while [ $((f+=1)) -le 8 ]
do
echo $f
done
Another way:
g=
while
g=${g}1
[ ${#g} -le 8 ]
do
echo ${#g}
done
and another:
set --
while
set $* .
[ ${#} -le 8 ]
do
echo ${#}
done
Here it worked on Mac OS X.
It includes the example of a BSD date, how to increment and decrement the date also:
for ((i=28; i>=6 ; i--));
do
dat=`date -v-${i}d -j "+%Y%m%d"`
echo $dat
done
Use:
max=10
for i in `eval echo {2..$max}`
do
echo $i
done
You need the explicit 'eval' call to reevaluate the {} after variable substitution.
doandthenkeywords on the same line asforandif, respectively. E.g.,for i in {2..10}; do