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divyashree

08-15-2013 10:49 PM

inserting variable in bash printf statement

In the following printf statement will print 1with padding 45 zeros.

Code:

printf '1%045d\n'

I want to put 45 in a variable(let n) and insert the($n) in the printf statement.

Can anyone help me to do it ?

Firerat

08-15-2013 11:07 PM

Not sure I understand

printf is doing what you asked,

Code:

for i in 0 {10..12};do

    printf '1%045d\n' $i

    printf '1% 45d\n' $i

    printf '1%-45dend\n' $i

    printf '1%045f\n' $i

done

1000000000000000000000000000000000000000000000

1                                            0

10                                            end

100000000000000000000000000000000000000.000000

1000000000000000000000000000000000000000000010

1                                          10

110                                          end

100000000000000000000000000000000000010.000000

1000000000000000000000000000000000000000000011

1                                          11

111                                          end

100000000000000000000000000000000000011.000000

1000000000000000000000000000000000000000000012

1                                          12

112                                          end

100000000000000000000000000000000000012.000000

if you post what you were expecting we can help you achieve that

divyashree

08-15-2013 11:19 PM

I want to replace the 45 in with a variable. Means I want like this

Quote:

printf '1%0$VARd\n'

SAbhi

08-15-2013 11:34 PM

to what i have understood it you can try something like this, though i would defer to @Firerat to comment on this further, he may have a better option.

Code:

printf 1%0${var}d

Firerat

08-15-2013 11:37 PM

Quote:

Originally Posted by divyashree (Post 5010152)

I want to replace the 45 in with a variable. Means I want like this

Sorry, you will have to give me more context

unless,,

Something like this?

Code:

for VAR in 0 {10..12};do

  printf '1%0'$VAR'd\n' $VAR

  printf '1% '$VAR'd\n' $VAR

  printf '1%-'$VAR'dend\n' $VAR

  printf '1%0'$VAR'f\n' $VAR

done

# gives you this

10

1 0

10end

10.000000

10000000010

1        10

110        end

1010.000000

100000000011

1        11

111        end

10011.000000

1000000000012

1          12

112          end

100012.000000

konsolebox

08-15-2013 11:41 PM

Place it also around double quotes to make it a little safer:

Code:

printf "%0${n}d\\n" 1

Firerat

08-15-2013 11:46 PM

Quote:

Originally Posted by konsolebox (Post 5010161)

Place it also around double quotes to make it a little safer:

Code:

printf "%0${n}d\\n" 1

yeah, I need to somehow commit the expansion order to memory.

NevemTeve

08-15-2013 11:50 PM

Code:

printf '1%0*d\n' "$WIDTH" "$VALUE"

divyashree

08-16-2013 12:36 AM

Quote:

Originally Posted by konsolebox (Post 5010161)

Place it also around double quotes to make it a little safer:

Code:

printf "%0${n}d\\n" 1

Thank you consolebox, I just simply forgot this substitution method.

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