Hi,
for i in 84 85 86 87 88 89 90 91 92; do ./runCleanup Session 11${i}; done
i tried as we do in perl like for i in 84..92; do ./runCleanup Session 11${i}; done
But its showing error.
what is the solution.
Thanks,
Deepak
Hi,
for i in 84 85 86 87 88 89 90 91 92; do ./runCleanup Session 11${i}; done
i tried as we do in perl like for i in 84..92; do ./runCleanup Session 11${i}; done
But its showing error.
what is the solution.
Thanks,
Deepak
set i=84
while [ $i <= 92 ]
do
# whatever
i=$(($i + 1))
done
or stick with the first version. What you don't seem to understand is that ".." is an operator (just like "+" is an operator) defined in Perl. It is, as you can see above, not available in Shell. It is a Perl language feature.
Hi masijade,
Thanks..
Hi,
for i in 84 85 86 87 88 89 90 91 92; do ./runCleanup Session 11${i}; done
i tried as we do in perl like for i in 84..92; do ./runCleanup Session 11${i}; done
But its showing error.what is the solution.
Thanks,
Deepak
You are just missing the expansion brackets.
for i in {84..92}; do
echo $i
done
Concerning the previous posted example
while [ $i <= 92 ]
Not a good idea. The `[' should not be use for arithmetic operation. `<=' are operators for strings. They might work; it will fail when least expected.
There is already a facility for that.
while (( $i <= 92 ))
do
echo $i
let ++i
done
Right, I believe however that simply using -lt instead of <= also works as expected.
Although I must say i did not know about {x .. y}, you learn something new everyday.
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