It's probably a lame question. But I am getting 3 arguments from command line [ bash script ]. Then I am trying to use these in a for loop.
for i in {$1..$2}
do act开发者_Python百科ion1
done
This doesn't seem to work though and if $1
is "0"
and $2
is 2
it prints {0..2}' and calls
action1` only once. I referred to various examples and this appears to be the correct usage. Can someone please tell me what needs to be fixed here?
How about:
for i in $(eval echo {$1..$2}); do echo $i; done
You can slice the input using ${@:3}
or ${@:3:8}
and then loop over it
For eg., to print arguments starting from 3
for i in ${@:3} ; do echo $i; done
or to print 8 arguments starting from 3 (so, arguments 3 through 10)
for i in ${@:3:8} ; do echo $i; done
Use the $@ variable?
for i in $@
do
echo $i
done
If you just want to use 1st and 2nd argument , just
for i in $1 $2
If your $1 and $2 are integers and you want to create a range, use the C for loop syntax (bash)
for ((i=$1;i<=$2;i++))
do
...
done
I had a similar problem. I think the issue is with dereferencing $1 within the braces '{}'. The following alternative worked for me ..
#!/bin/bash
for ((i=$1;i<=$2;i++))
do
...
done
Hope that helps.
for i in `seq $1 $2`; do echo $i; done
#/bin/bash
for i
do
echo Value: $i
done
This will loop over all arguments given to the script file. Note, no "do" or anything else after the loop variable i.
I recommend this: In loop we handle only $1, and do shift N times (if arguments will be need next, then better save it, because shift removing argument):
for a in 1 2 3
do
echo $1
shift
done
Example: ./test.sh "abra k" "b 2" "c 14"
abra k
b 2
c 14
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