How to concatenate string variables in Bash

Question

In PHP, strings are concatenated together as follows:

$foo = \"Hello\";
$foo .= \" World\";

Here, $foo becomes \"Hello World\"

Answer 1

Bash first

As this question stand specifically for Bash, my first part of the answer would present different ways of doing this properly:

+=: Append to variable

The syntax += may be used in different ways:

Append to string var+=...

(Because I am frugal, I will only use two variables foo and a and then re-use the same in the whole answer. ;-)

a=2
a+=4
echo $a
24

Using the Stack Overflow question syntax,

foo="Hello"
foo+=" World"
echo $foo
Hello World

works fine!

Append to an integer ((var+=...))

variable a is a string, but also an integer

echo $a
24
((a+=12))
echo $a
36

Append to an array var+=(...)

Our a is also an array of only one element.

echo ${a[@]}
36

a+=(18)

echo ${a[@]}
36 18
echo ${a[0]}
36
echo ${a[1]}
18

Note that between parentheses, there is a space separated array. If you want to store a string containing spaces in your array, you have to enclose them:

a+=(one word "hello world!" )
bash: !": event not found

Hmm.. this is not a bug, but a feature... To prevent bash to try to develop !", you could:

a+=(one word "hello world"! 'hello world!' $'hello world\041')

declare -p a
declare -a a='([0]="36" [1]="18" [2]="one" [3]="word" [4]="hello world!" [5]="h
ello world!" [6]="hello world!")'

printf: Re-construct variable using the builtin command

The printf builtin command gives a powerful way of drawing string format. As this is a Bash builtin, there is a option for sending formatted string to a variable instead of printing on stdout:

echo ${a[@]}
36 18 one word hello world! hello world! hello world!

There are seven strings in this array. So we could build a formatted string containing exactly seven positional arguments:

printf -v a "%s./.%s...'%s' '%s', '%s'=='%s'=='%s'" "${a[@]}"
echo $a
36./.18...'one' 'word', 'hello world!'=='hello world!'=='hello world!'

Or we could use one argument format string which will be repeated as many argument submitted...

Note that our a is still an array! Only first element is changed!

declare -p a
declare -a a='([0]="36./.18...'\''one'\'' '\''word'\'', '\''hello world!'\''=='\
''hello world!'\''=='\''hello world!'\''" [1]="18" [2]="one" [3]="word" [4]="hel
lo world!" [5]="hello world!" [6]="hello world!")'

Under bash, when you access a variable name without specifying index, you always address first element only!

So to retrieve our seven field array, we only need to re-set 1st element:

a=36
declare -p a
declare -a a='([0]="36" [1]="18" [2]="one" [3]="word" [4]="hello world!" [5]="he
llo world!" [6]="hello world!")'

One argument format string with many argument passed to:

printf -v a[0] '<%s>\n' "${a[@]}"
echo "$a"
<36>
<18>

Using the Stack Overflow question syntax:

foo="Hello"
printf -v foo "%s World" $foo
echo $foo
Hello World

Nota: The use of double-quotes may be useful for manipulating strings that contain spaces, tabulations and/or newlines

printf -v foo "%s World" "$foo"

Shell now

Under POSIX shell, you could not use bashisms, so there is no builtin printf.

Basically

But you could simply do:

foo="Hello"
foo="$foo World"
echo $foo
Hello World

Formatted, using forked printf

If you want to use more sophisticated constructions you have to use a fork (new child process that make the job and return the result via stdout):

foo="Hello"
foo=$(printf "%s World" "$foo")
echo $foo
Hello World

Historically, you could use backticks for retrieving result of a fork:

foo="Hello"
foo=`printf "%s World" "$foo"`
echo $foo
Hello World

But this is not easy for nesting:

foo="Today is: "
foo=$(printf "%s %s" "$foo" "$(date)")
echo $foo
Today is: Sun Aug 4 11:58:23 CEST 2013

with backticks, you have to escape inner forks with backslashes:

foo="Today is: "
foo=`printf "%s %s" "$foo" "\`date\`"`
echo $foo
Today is: Sun Aug 4 11:59:10 CEST 2013