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.

didn't work yet. it prints all elements and then print a new line — milad
– milad, Commented Feb 22, 2019 at 8:21

Rahul · Accepted Answer · 2019-02-22 09:04:20Z

1

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

echo -n ${li[$j]}

Check docs here.

answered Feb 22, 2019 at 8:58

Rahul

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1 Comment

didn't work, i want to have new line after any word but it doesn't do that

l.kusiak · Accepted Answer · 2019-02-22 08:40:23Z

0

Try with this:

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

answered Feb 22, 2019 at 8:40

l.kusiak

162 bronze badges

it works fine, thank you. but i have a question: why i can't do that by echo?