how to print two array variables in a line in BASH

Question

I tried this code in my shell script in ubuntu 12.04 bash

IFS=$'\t'
name=(one two three four five)
fam=($(seq -s" " 1 1 5))
for (i=0;i<5;i++) 
do 
  printf "%s\t%s\n" ${fam[i]} ${name[i]} 
done

The output I want is like that

1 one
2 two
3 three
4 four
5 five

But the really output is

1 2 3 4 5   one
two 
three   
four    
five

What went wrong in my code? and how to print more than one array variable in a line just using one loop?

Zombo · Accepted Answer · 2013-06-20 07:56:34Z

2

name=(one two three four five)
fam=(1 2 3 4 5)

for i in ${!name[*]}
do
  printf '%s %s\n' ${fam[i]} ${name[i]}
done

or you can just fix

for  (i = 0; i < 5; i++)
for ((i = 0; i < 5; i++))

answered Jun 20, 2013 at 7:56

Zombo

1

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1 Comment

Yao He Over a year ago

Because if I want to get a big sequence like 10000,it is not a good way input all numbers by keyboard?What do you think?

ccarton · Accepted Answer · 2013-06-20 11:31:41Z

0

When you set IFS to '\t' it causes the array initializer to break the input on tabs so you get '1 2 3 4 5' assigned to fam[0]. You need to change the seq delimiter to match.

fam=($(seq -s $'\t' 1 1 5))

answered Jun 20, 2013 at 11:31

ccarton

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