I'm going to presume that you want to search for a specific set of numbers, and that the fact that the fact that they're odd doesn't matter....
In that case, you'd have to use a for loop to seach thru the list of 'important' values:
for $tval in 1 3 5 7 9; do if [ $a -eq $tval ] then echo -n "$a " fi done
As far as I know, there's no fast way to search for one value in a list of candidates.
Scott@Charter wrote:
I can't get this bash script to work. It's suppose to print all odd numbers from 1 to 10.
#!/usr/bin/bash
LIMIT=10 a=1
while [ "$a" -le $LIMIT ] do if [ "$a" -eq $(1 3 5 7 9) ] <---------Something not right here. then echo -n "$a " fi
echo
let "a+=1" done echo; echo exit 0
Can anyone figure out what I am doing wrong? Can I also see the same thing written in Perl?
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