formatting the output using printf in bash script

Question

I am writing a script to display a menu at the center of the screen for the users to select the options using printf command in bash script.

I am finding the middle column of the screen, and start printing the messages from the middle column. Hence, the output will be displayed at the center.

Below is the code snippet

#!/bin/bash

INSTALLATION_HEADER=" Installater Options "

COLS=$(tput cols)

print_header ()
{
        local equal=()
        local title="$1"
        local mid=$(((${#title}+$COLS)/2))
        for (( i=0;  $i < ${#title} ; i=$(($i+1)) ))
        do
                local hypen+="-"
        done
        printf "%*s\n" $mid "$title"
        printf "%*s\n" $mid "$hypen"
        echo ""
        echo ""
}

print_install_options ()
{
        local title=${1}
        local mid=$(((${#title}+$COLS)/2))
        for (( i=0;  $i < ${#title} ; i=$(($i+1)) ))
        do
                local hypen+="-"
        done

        printf "%*s\n" $mid "$hypen"
        for i in "${install_options[@]}" ;
        do
                printf "%*s\n" $mid "$i"
        done
        printf "%*s\n" $mid "$hypen"
}

install_options=("1. Install" "2. Uninstall")
print_header "$INSTALLATION_HEADER"
print_install_options "$INSTALLATION_HEADER"

When I execute the above code, the output produced is

 Installater Options
---------------------


---------------------
           1. Install
         2. Uninstall
---------------------

The expected output should be

 Installater Options
---------------------


---------------------
1. Install
2. Uninstall
---------------------

"1. Ïnstall" and "2. Uninstall" are not printing at the middle of the screen. Please help me in resolving it.

Thanks in Advance.

Updated Script:

Thanks to all for anwering my question.

Below is the script which gives the required output.

#!/bin/bash

INSTALLATION_HEADER=" Installater Options "

COLS=$(tput cols)

print_header ()
{
        local equal=()
        local title="$1"
        local mid=$(((${#title}+$COLS)/2))
        for (( i=0;  $i < ${#title} ; i=$(($i+1)) ))
        do
                local hypen+="-"
        done
        printf "%*s\n" $mid "$title"
        printf "%*s\n" $mid "$hypen"
        echo ""
        echo ""
}

print_install_options ()
{
        local title=${1}
        local length=$(((${#title}+$COLS)))
        local mid=$(((${#title}+$COLS)/2))
        for (( i=0;  $i < ${#title} ; i=$(($i+1)) ))
        do
                local hypen+="-"
        done

        printf "%*s\n" $mid "$hypen"
        for i in "${install_options[@]}" ;
        do
                printf "%*s%s" $((${mid}-${#title})) "" "|"
                printf "%s" "  $i  "
                printf "%*s\n" $((${#title}-${#i}-5)) "|"
        done
        printf "%*s\n" $mid "$hypen"
}

install_options=("1. Install" "2. Uninstall")
print_header "$INSTALLATION_HEADER"
print_install_options "$INSTALLATION_HEADER"

Output:

 Installater Options
---------------------


---------------------
|  1. Install       |
|  2. Uninstall     |
---------------------

@sach I have added an answer where I address, in a note, the problem you might have referencing $COLUMNS inside a script. Resumé: $COLUMNS works if you source the script. — gboffi
– gboffi, Commented Nov 13, 2014 at 14:21

user3620917 · Accepted Answer · 2014-11-13 14:14:08Z

3

Change line 34 to

printf "%-*s%s\n" $((${mid}-${#title})) " " "$i"

Result:

                                 Installater Options 
                                ---------------------


                                ---------------------
                                1. Install
                                2. Uninstall
                                ---------------------

answered Nov 13, 2014 at 14:14

user3620917

Sign up to request clarification or add additional context in comments.

3 Comments

Nikhil Over a year ago

It works. Could you please explain the printf used here.

user3620917 Over a year ago

@sach $((${mid}-${#title})) is arithmetic operation which result is $mid minus length of title. So we printf two string (%s): first one is just bare space " " aligned to the left and repeated mid minus #title times. The second string is $i.

Nikhil Over a year ago

Thanks rooom, I have updated my script in the question, please have a look. Any changes plz suggest.

John Zwinck · Accepted Answer · 2014-11-13 13:51:29Z

1

Instead of this:

            printf "%*s\n" $mid "$i"

You need this:

            printf "%-*s\n" $mid "$i"

The - means left-align.

answered Nov 13, 2014 at 13:51

John Zwinck

252k44 gold badges346 silver badges459 bronze badges

3 Comments

Nikhil Over a year ago

I am trying to print the statement which should start at the mid ( which is the center of the screen) value. Here left alignment does not work.

user3442743 Over a year ago

@sach the middle of the screen or the middle of the dashed lines ?

Nikhil Over a year ago

@Jidder, Its middle of the screen.

gboffi · Accepted Answer · 2014-11-13 14:11:17Z

1

The question is about printing in the centre of a screen, even if the examples refer to a left aligned string (see John's answer for that...) so I will give the answer for centering a string even if it is of tangential interest for the OP

centre () {
    nc=`tput cols`; l=${#1}
    printf "%$(( (nc-l)/2 ))s%s\n" " " "$1"
    }

Comment for the OP

In your code you reference $COLUMNS, look at this:

% cat aaaa
echo pippo $COLUMNS
tput cols
% source aaaa
pippo 117
117
% sh aaaa
pippo
117
%

and use tput to find the number of columns

answered Nov 13, 2014 at 14:11

gboffi

25.4k10 gold badges62 silver badges98 bronze badges

Collectives™ on Stack Overflow

formatting the output using printf in bash script

Updated Script:

3 Answers 3

3 Comments

3 Comments

Comment for the OP

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

Updated Script:

3 Answers 3

3 Comments

3 Comments

Comment for the OP

Comments

Your Answer

Sign up or log in

Post as a guest

Related