In PHP, strings are concatenated together as follows:

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

Here, $foo becomes "Hello World".

How is this accomplished in Bash?

---

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

一般来说，要连接两个变量，你可以一个接一个地写出它们：

a='hello'
b='world'
c=$a$b
echo $c
> helloworld

---

Bash还支持+ =运算符，如下面的记录所示：

$ A="X Y"
$ A+="Z"
$ echo "$A"
X YZ

---

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 world41')

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 formated 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 formated 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 wich will be repeated as many argument submited...

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>
<one>
<word>
<hello world!>
<hello world!>
<hello world!>

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

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