accessing elements in array in script

Question

i am trying to print all parameters in bash script "one by one".
fro example i want to run:
./myscript hello all friends
and see below result:

 hello  
 all  
 friends.

i wrote below code:

#!/bin/bash
li=$@
for(( j=0;j<$#;j++));
do
    echo ${li[$j]}
done

bug when i run my code it prints all of argument at once:

hello all friends

i know i can do that by changing the for structure to below format:

#!/bin/bash
li=$@
for j in $li;
do
    echo $j
done

but i didn't want to change the code such as above.
please help me.
thank you in advance.

https://stackoverflow.com/questions/8467424/echo-newline-in-bash-prints-literal-n — l.kusiak, Feb 22 '19 at 08:19
didn't work yet. it prints all elements and then print a new line — milad, Feb 22 '19 at 08:21

Rahul · Answer 1 · 2019-02-22T09:04:20.793

1

You can write using echo -n option to skip printing newline at the end.

echo -n ${li[$j]}

Check docs here.

edited Feb 22 '19 at 09:04

answered Feb 22 '19 at 08:58

Rahul

didn't work, i want to have new line after any word but it doesn't do that – milad Feb 22 '19 at 12:31

score 0 · Accepted Answer · answered Feb 22 '19 at 08:40

0

Try with this:

#!/bin/bash
li=$@
for(( j=0;j<$#;j++)); do
    printf '%s\n' ${li[$j]}
done

answered Feb 22 '19 at 08:40

l.kusiak

it works fine, thank you. but i have a question: why i can't do that by echo? – milad Feb 22 '19 at 12:40

2 Answers2