I have two variables var1 and var2. The contents of each variables come from bash shell grep command.
echo $var1 prints
123 465 326 8080
echo $var2 prints
sila kiran hinal juku
Now I want to print the above into following formats in Linux bash shell
123 sila 465 kiran 326 hinal 8080 juku
So how can I print this way in bash shell??
What about?
$ paste -d" " <(echo $var1 | xargs -n1) <(echo $var2 | xargs -n1)
We can even skip the echo:
$ paste -d" " <(xargs -n1 <<< $var1) <(xargs -n1 <<< $var2)
Without a loop:
$ var1="123 465 326 8080"
$ var2="sila kiran hinal juku"
$ var1=($var1); var2=($var2)
$ saveIFS=IFS
$ IFS=$'\n'
$ paste <(echo "${a[*]}") <(echo "${b[*]}"
$ IFS=$saveIFS
With a loop (assumes that the two strings have the same number of words):
$ var1="123 465 326 8080"
$ var2="sila kiran hinal juku"
$ var2=($var2)
$ for s in $var1; do echo $s ${vars[i++]}; done
Using file descriptors and a while-loop:
var1="123 465 326 8080"
var2="sila kiran hinal juku"
IFS=" " exec 7< <(printf "%s\n" $var1) 8< <(printf "%s\n" $var2)
while read -u7 f1 && read -u8 f2; do
echo "$f1 $f2"
done
I'd store them into arrays $var1[] and $var2[] instead of a some long string and then iterate through the arrays with a loop to output it the way you want.
If you don't want ot use arrays, you could use awk and the iterator from a loop to print out the names one at a time.
join <(echo $var1 | sed -r 's/ +/\n/g' | cat -n) <(echo $var2 | sed -r 's/ +/\n/g' | cat -n) -o "1.2,2.2"
joincommand, except I'd need to see how $var1 and $var2 got created.