Printing multiple words in a line using printf() in bash

Question

Im trying to print like below

2333 k4k3 34kk 34r4
233k 2344 234l llll

with command

printf "%04d \n" "1111" "1123" "3333" "3333"

but its breaks it. what I want to happen is all 4 values will be in single line and then add new line again.

Inian · Accepted Answer · 2017-11-15 07:48:03Z

1

The C style printf in bash expects each format specifier and an associated argument to be provided. For example

printf "Surname: %s Name: %s\n" "foo" "bar"

So you need to provide the specifiers for each of the arguments needed, something like

printf "%04d %04d %04d %04d\n" "1111" "1123" "3333" "3333"

(or) if you want to treat the entire set of words as a single string literal, you just need to use a single string format specifier. Notice below the quotes covering the entire set of arguments as one.

printf "%s \n" "1111 1123 3333 3333"

See The printf command in bash for more options and help articles.

edited Nov 15, 2017 at 7:48

answered Nov 15, 2017 at 7:41

Inian

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1 Comment

error467 Over a year ago

works fine. I guess it needs to set identifier for every argument. thanks lnian

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