Concatenating multiple line variables using newline

Question

I have two variables var1 and var2 that could either be multiple lines or could be completely empty. For example, either one could look like:

line/1
line/2

or

I want to concatenate them so that when put together var1 outputs as is, and var 2 outputs as is right below the last line of var1. There are 4 cases, either var1 is empty and var2 is not, var 2 is empty and var 1 is not, both are empty, or neither are empty. I dont want any whitespace or an empty line if one of the variables is empty. So if var 1 is empty I do not want,


line/1

or vice versa.

Other than using a if, elif, else block, is there a way I could do this or do I HAVE to use an if else block.

In addition, for the last case where neither are empty how can I concatenate these two? I have tried

var3="${var1}\n${var2}"

but that doesn't seem to work. Any tips would be much appreciated.

\n does not interpolate a newline character, but a literal ` followed by a n. You can verify this by doing a, say, xxd <<<$var3`. — user1934428
– user1934428, Commented Apr 20, 2020 at 6:20

anubhava · Accepted Answer · 2020-04-20 06:32:19Z

2

Using bash, you can use this function for this job:

concat() {
   printf '%s' "${1:+$1$'\n'}" "${2:+$2$'\n'}";
}

Use of "${1:+$1$'\n'}" appends \n to $1 only if $1 is null/unset.

As per man bash:

${parameter:+word} Use Alternate Value. If parameter is null or unset, nothing is substituted, otherwise the expansion of word is substituted.

Examples:

concat 'line1' 'line2'

line1
line2

concat 'line1' ''

line1

concat '' 'line2'

line2

concat '' ''

concat 'line1
line2
line3' ''

line1
line2
line3

edited Apr 20, 2020 at 6:32

answered Apr 20, 2020 at 6:24

anubhava

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1 Comment

JavaIsOkay1170 Over a year ago

thank you, this actually worked perfectly. The explanation was very insightful as well!

KamilCuk · Accepted Answer · 2020-04-20 06:37:47Z

2

var3="$var1${var1:+${var2:+$'\n'}}$var2"

should work.

$ var1=; var2=; var3="${var1}${var1:+${var2:+$'\n'}}${var2}"; echo "$var3"
               # empty line from echo
$ var1=1; var2=; var3="${var1}${var1:+${var2:+$'\n'}}${var2}"; echo "$var3"
1
$ var1=; var2=2; var3="${var1}${var1:+${var2:+$'\n'}}${var2}"; echo "$var3"
2
$ var1=1; var2=2; var3="${var1}${var1:+${var2:+$'\n'}}${var2}"; echo "$var3"
1
2

${var1} - put var1
${var1:+ - when var1 is set and
- ${var2:+ - when var2 is set
  - $'\n' - then add a newline
- }
}
${var2} - and add var2 too.

Basically the ${var1:+${var2:+$'\n'}} adds a newline only if var1 and var2 are both set, to separate var1 and var2.

For reference bash manual shell parameter expansion.

edited Apr 20, 2020 at 6:37

answered Apr 20, 2020 at 6:29

KamilCuk

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1 Comment

KamilCuk Over a year ago

It's literally explained below. Maybe refresh the page, I edited the post.

Jetchisel · Accepted Answer · 2020-04-20 06:19:53Z

1

You can try using printf -v

printf -v var3 '%s\n "$var1" "$var2"

answered Apr 20, 2020 at 6:19

Jetchisel

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4 Comments

JavaIsOkay1170 Over a year ago

what exactly does the -v modifier do? What does the %s\n do?

KamilCuk Over a year ago

Still, when var1 or var2 are empty, this will result in empty lines.

Jetchisel Over a year ago

@KamilCuk indeed

Jetchisel Over a year ago

@nutella100 help prinf -v var assign the output to shell variable VAR rather than display it on the standard output

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