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Loop iteration with two variables

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# 8  
Old 07-30-2016
This is tested in bash, but a recent ksh offers the central functionalities as well:
Code:
{ exec 4<&0; { exec 3<&0; while read -u3 -d" " A && read -u4 -d" " B; do echo $A $B; done; } <<< $group1" "; } <<< $group2" "
Joe Kim
Brian Annie
John Kate
Mark Anna


EDIT: Actually, the file descriptor 3 stuff can be omitted and the normal stdin be used.
# 9  
Old 07-30-2016
Quote:
Originally Posted by RudiC
ksh allows for arrays, so this might help:
Code:
A1=($group1)
A2=($group2)
for i in $(seq 0 ${#A1[@]}); do echo ${A1[$i]} ${A2[$i]}; done
Joe Kim
Brian Annie
John Kate
Mark Anna

Might need some polishing, as there's an trailing empty line...


EDIT: Hola, Don Cragun just outperformed me by 2 minutes...
You can get rid of the trailing empty line by changing ${#A1[@]} in the above to $((${#A1[@]}-1)). I usually avoid seq in loops like this because it isn't a built-in in either ksh or bash (at least on OS X).
# 10  
Old 07-30-2016
In bash, ${!A1[@]} returns all the indices of the array A1, so this simplified approach will work:

Code:
for i in ${!A1[@]}; do echo ${A1[$i]} ${A2[$i]}; done
Joe Kim
Brian Annie
John Kate
Mark Anna

Not sure if it does in ksh, though...
# 11  
Old 07-30-2016
Quote:
Originally Posted by RudiC
In bash, ${!A1[@]} returns all the indices of the array A1, so this simplified approach will work:

Code:
for i in ${!A1[@]}; do echo ${A1[$i]} ${A2[$i]}; done
Joe Kim
Brian Annie
John Kate
Mark Anna

Not sure if it does in ksh, though...
${!array[@]} expands to a list of the subscripts of the current elements of array in 1993 and later versions ksh. In 1988 versions of ksh it is a syntax error.
This User Gave Thanks to Don Cragun For This Post:
Decoy Octopus88
# 12  
Old 07-30-2016
With standard Posix stuff:
Code:
group1="Joe Brian John Mark"
group2="Kim Annie Kate Anna"
exec 3<<EOF
$(printf "%s\n" $group2)
EOF
while read name1 && read name2 <&3
do
  echo "$name1:$name2"
done <<EOF
$(printf "%s\n" $group1)
EOF

Or, if you use newlines as list separators (which is better if you use names with spaces, plus it saves two subshells):
Code:
group1="\
Joe
Brian
John
Mark"

group2="\
Kim
Annie
Kate
Anna"

exec 3<<EOF
$group2
EOF

while read name1 && read name2 <&3
do
  echo "$name1:$name2"
done <<EOF
$group1
EOF

Output:
Code:
Joe:Kim
Brian:Annie
John:Kate
Mark:Anna

This User Gave Thanks to Scrutinizer For This Post:
Decoy Octopus88
# 13  
Old 07-30-2016
Rudy and Don. I really appreciate your responses, and I was able to get both these to run on ksh93 my original diff -r tasks. I hate to flip flop on you, but i was using the print names as an easy example to follow. However it is not running on ksh88. To cut to the chase, i am able to get below code to run ksh88 as long as there is only 1 directory per group, however I need it to run with multiple directories. I will post both examples below. If you have any ideas please let me know. Scrutinizer i will have a look at your example next. Thanks again.

This below does not work in ksh88 when there are two directories per group 1/2.
Code:
#/usr/bin/ksh
group1="/usr/bin /usr/games"
group2="/usr/local/bin /usr/local/games"
i=0
j=0
while [ $i -lt ${#group1[@]} ] && [ $j -lt ${#group2[@]} ]
do	diff -r "${group1[i]}" "${group2[j]}"
	i=$((i + 1))
        j=$((j + 1)) 
done

This below WILL work in ksh88 with only one directory variable per group, note i cannot use () to surround the group1/2 variables in ksh88.

Code:
#/usr/bin/ksh
group1="/usr/bin"
group2="/usr/local/bin"
i=0
j=0
while [ $i -lt ${#group1[@]} ] && [ $j -lt ${#group2[@]} ]
do	diff -r "${group1[i]}" "${group2[j]}"
	i=$((i + 1))
        j=$((j + 1)) 
done

# 14  
Old 07-31-2016
Please show us the exact output you get from running the following script:
Code:
#/usr/bin/ksh
set -xv
group1=("/usr/bin" "/usr/games")
group2=("/usr/local/bin" "/usr/local/games")
i=0
while [ $i -lt ${#group1[@]} ] && [ $i -lt ${#group2[@]} ]
do	echo diff -r "${group1[i]}" "${group2[i]}"
	i=$((i + 1))
done

This User Gave Thanks to Don Cragun For This Post:
Decoy Octopus88
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