In PHP, strings are concatenated together as follows:
$foo = "Hello";
$foo .= " World";
Here, $foo becomes "Hello World".
How is this accomplished in Bash?
반응형
foo="Hello"
foo="${foo} World"
echo "${foo}"
> Hello World
In general to concatenate two variables you can just write them one after another:
a='Hello'
b='World'
c="${a} ${b}"
echo "${c}"
> Hello World
codaddict
Bash also supports a += operator as shown in this code:
A="X Y"
A+=" Z"
echo "$A"
output
X Y Z
thkala
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>
<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
F. Hauri
You can do this too:
$ var="myscript"
$ echo $var
myscript
$ var=${var}.sh
$ echo $var
myscript.sh
userend
bla=hello
laber=kthx
echo "${bla}ohai${laber}bye"
Will output
helloohaikthxbye
This is useful when
$blaohai
leads to a variable not found error. Or if you have spaces or other special characters in your strings. "${foo}" properly escapes anything you put into it.
orkoden
foo="Hello "
foo="$foo World"
vinothkr
Here is a concise summary of what most answers are talking about.
Let's say we have two variables and $1 is set to 'one':
set one two
a=hello
b=world
The table below explains the different contexts where we can combine the values of a and b to create a new variable, c.
Context | Expression | Result (value of c)
--------------------------------------+-----------------------+---------------------
Two variables | c=$a$b | helloworld
A variable and a literal | c=${a}_world | hello_world
A variable and a literal | c=$1world | oneworld
A variable and a literal | c=$a/world | hello/world
A variable, a literal, with a space | c=${a}" world" | hello world
A more complex expression | c="${a}_one|${b}_2" | hello_one|world_2
Using += operator (Bash 3.1 or later) | c=$a; c+=$b | helloworld
Append literal with += | c=$a; c+=" world" | hello world
A few notes:
- enclosing the RHS of an assignment in double quotes is generally a good practice, though it is quite optional in many cases
+=is better from a performance standpoint if a big string is being constructed in small increments, especially in a loop- use
{}around variable names to disambiguate their expansion (as in row 2 in the table above). As seen on rows 3 and 4, there is no need for{}unless a variable is being concatenated with a string that starts with a character that is a valid first character in shell variable name, that is alphabet or underscore.
See also:
codeforester
The way I'd solve the problem is just
$a$b
For example,
a="Hello"
b=" World"
c=$a$b
echo "$c"
which produces
Hello World
If you try to concatenate a string with another string, for example,
a="Hello"
c="$a World"
then echo "$c" will produce
Hello World
with an extra space.
$aWorld
doesn't work, as you may imagine, but
${a}World
produces
HelloWorld
Chris Smith
$ a=hip
$ b=hop
$ ab=$a$b
$ echo $ab
hiphop
$ echo $a$b
hiphop
bcosca
Yet another approach...
> H="Hello "
> U="$H""universe."
> echo $U
Hello universe.
...and yet yet another one.
> H="Hello "
> U=$H"universe."
> echo $U
Hello universe.
Akseli Palén
If you want to append something like an underscore, use escape (\)
FILEPATH=/opt/myfile
This does not work:
echo $FILEPATH_$DATEX
This works fine:
echo $FILEPATH\\_$DATEX
user2800471
The simplest way with quotation marks:
B=Bar
b=bar
var="$B""$b""a"
echo "Hello ""$var"
betontalpfa
Even if the += operator is now permitted, it has been introduced in Bash 3.1 in 2004.
Any script using this operator on older Bash versions will fail with a "command not found" error if you are lucky, or a "syntax error near unexpected token".
For those who cares about backward compatibility, stick with the older standard Bash concatenation methods, like those mentioned in the chosen answer:
foo="Hello"
foo="$foo World"
echo $foo
> Hello World
Louis-Félix
You can concatenate without the quotes. Here is an example:
$Variable1 Open
$Variable2 Systems
$Variable3 $Variable1$Variable2
$echo $Variable3
This last statement would print "OpenSystems" (without quotes).
This is an example of a Bash script:
v1=hello
v2=world
v3="$v1 $v2"
echo $v3 # Output: hello world
echo "$v3" # Output: hello world
mariana soffer
I prefer to use curly brackets ${} for expanding variable in string:
foo="Hello"
foo="${foo} World"
echo $foo
> Hello World
Curly brackets will fit to Continuous string usage:
foo="Hello"
foo="${foo}World"
echo $foo
> HelloWorld
Otherwise using foo = "$fooWorld" will not work.
Nick Tsai
If what you are trying to do is to split a string into several lines, you can use a backslash:
$ a="hello\
> world"
$ echo $a
helloworld
With one space in between:
$ a="hello \
> world"
$ echo $a
hello world
This one also adds only one space in between:
$ a="hello \
> world"
$ echo $a
hello world
jcarballo
Safer way:
a="AAAAAAAAAAAA"
b="BBBBBBBBBBBB"
c="CCCCCCCCCCCC"
d="DD DD"
s="${a}${b}${c}${d}"
echo "$s"
AAAAAAAAAAAABBBBBBBBBBBBCCCCCCCCCCCCDD DD
Strings containing spaces can become part of command, use "$XXX" and "${XXX}" to avoid these errors.
Plus take a look at other answer about +=
Bohdan
There's one particular case where you should take care:
user=daniel
cat > output.file << EOF
"$user"san
EOF
Will output "daniel"san, and not danielsan, as you might have wanted.
In this case you should do instead:
user=daniel
cat > output.file << EOF
${user}san
EOF
diogovk
Despite of the special operator, +=, for concatenation, there is a simpler way to go:
foo='Hello'
foo=$foo' World'
echo $foo
Double quotes take an extra calculation time for interpretation of variables inside. Avoid it if possible.
Tiys
a="Hello,"
a=$a" World!"
echo $a
This is how you concatenate two strings.
CodeNinjaPI
If it is as your example of adding " World" to the original string, then it can be:
#!/bin/bash
foo="Hello"
foo=$foo" World"
echo $foo
The output:
Hello World
nonopolarity
var1='hello'
var2='world'
var3=$var1" "$var2
echo $var3
hari
There are voiced concerns about performance, but no data is offered. Let me suggest a simple test.
(NOTE: date on macOS does not offer nanoseconds, so this must be done on Linux.)
I have created append_test.sh on GitHub with the contents:
#!/bin/bash -e
output(){
ptime=$ctime;
ctime=$(date +%s.%N);
delta=$(bc <<<"$ctime - $ptime");
printf "%2s. %16s chars time: %s delta: %s\n" $n "$(bc <<<"10*(2^$n)")" $ctime $delta;
}
method1(){
echo 'Method: a="$a$a"'
for n in {1..32}; do a="$a$a"; output; done
}
method2(){
echo 'Method: a+="$a"'
for n in {1..32}; do a+="$a"; output; done
}
ctime=0; a="0123456789"; time method$1
Test 1:
$ ./append_test.sh 1
Method: a="$a$a"
1. 20 chars time: 1513640431.861671143 delta: 1513640431.861671143
2. 40 chars time: 1513640431.865036344 delta: .003365201
3. 80 chars time: 1513640431.868200952 delta: .003164608
4. 160 chars time: 1513640431.871273553 delta: .003072601
5. 320 chars time: 1513640431.874358253 delta: .003084700
6. 640 chars time: 1513640431.877454625 delta: .003096372
7. 1280 chars time: 1513640431.880551786 delta: .003097161
8. 2560 chars time: 1513640431.883652169 delta: .003100383
9. 5120 chars time: 1513640431.886777451 delta: .003125282
10. 10240 chars time: 1513640431.890066444 delta: .003288993
11. 20480 chars time: 1513640431.893488326 delta: .003421882
12. 40960 chars time: 1513640431.897273327 delta: .003785001
13. 81920 chars time: 1513640431.901740563 delta: .004467236
14. 163840 chars time: 1513640431.907592388 delta: .005851825
15. 327680 chars time: 1513640431.916233664 delta: .008641276
16. 655360 chars time: 1513640431.930577599 delta: .014343935
17. 1310720 chars time: 1513640431.954343112 delta: .023765513
18. 2621440 chars time: 1513640431.999438581 delta: .045095469
19. 5242880 chars time: 1513640432.086792464 delta: .087353883
20. 10485760 chars time: 1513640432.278492932 delta: .191700468
21. 20971520 chars time: 1513640432.672274631 delta: .393781699
22. 41943040 chars time: 1513640433.456406517 delta: .784131886
23. 83886080 chars time: 1513640435.012385162 delta: 1.555978645
24. 167772160 chars time: 1513640438.103865613 delta: 3.091480451
25. 335544320 chars time: 1513640444.267009677 delta: 6.163144064
./append_test.sh: fork: Cannot allocate memory
Test 2:
$ ./append_test.sh 2
Method: a+="$a"
1. 20 chars time: 1513640473.460480052 delta: 1513640473.460480052
2. 40 chars time: 1513640473.463738638 delta: .003258586
3. 80 chars time: 1513640473.466868613 delta: .003129975
4. 160 chars time: 1513640473.469948300 delta: .003079687
5. 320 chars time: 1513640473.473001255 delta: .003052955
6. 640 chars time: 1513640473.476086165 delta: .003084910
7. 1280 chars time: 1513640473.479196664 delta: .003110499
8. 2560 chars time: 1513640473.482355769 delta: .003159105
9. 5120 chars time: 1513640473.485495401 delta: .003139632
10. 10240 chars time: 1513640473.488655040 delta: .003159639
11. 20480 chars time: 1513640473.491946159 delta: .003291119
12. 40960 chars time: 1513640473.495354094 delta: .003407935
13. 81920 chars time: 1513640473.499138230 delta: .003784136
14. 163840 chars time: 1513640473.503646917 delta: .004508687
15. 327680 chars time: 1513640473.509647651 delta: .006000734
16. 655360 chars time: 1513640473.518517787 delta: .008870136
17. 1310720 chars time: 1513640473.533228130 delta: .014710343
18. 2621440 chars time: 1513640473.560111613 delta: .026883483
19. 5242880 chars time: 1513640473.606959569 delta: .046847956
20. 10485760 chars time: 1513640473.699051712 delta: .092092143
21. 20971520 chars time: 1513640473.898097661 delta: .199045949
22. 41943040 chars time: 1513640474.299620758 delta: .401523097
23. 83886080 chars time: 1513640475.092311556 delta: .792690798
24. 167772160 chars time: 1513640476.660698221 delta: 1.568386665
25. 335544320 chars time: 1513640479.776806227 delta: 3.116108006
./append_test.sh: fork: Cannot allocate memory
The errors indicate that my Bash got up to 335.54432 MB before it crashed. You could change the code from doubling the data to appending a constant to get a more granular graph and failure point. But I think this should give you enough information to decide whether you care. Personally, below 100 MB I don't. Your mileage may vary.
Bruno Bronosky
I wanted to build a string from a list. Couldn't find an answer for that so I post it here. Here is what I did:
list=(1 2 3 4 5)
string=''
for elm in "${list[@]}"; do
string="${string} ${elm}"
done
echo ${string}
and then I get the following output:
1 2 3 4 5
simibac
Note that this won't work
foo=HELLO
bar=WORLD
foobar=PREFIX_$foo_$bar
as it seems to drop $foo and leaves you with:
PREFIX_WORLD
but this will work:
foobar=PREFIX_"$foo"_"$bar"
and leave you with the correct output:
PREFIX_HELLO_WORLD
Dss
Here is the one through AWK:
$ foo="Hello"
$ foo=$(awk -v var=$foo 'BEGIN{print var" World"}')
$ echo $foo
Hello World
Avinash Raj
I do it this way when convenient: Use an inline command!
echo "The current time is `date`"
echo "Current User: `echo $USER`"
Marty
Variables and arrays (indexed or associative*) in bash are always strings by default, but you can use flags to the declare builtin, to give them attributes like "integer" (-i) or "reference"** (-n), which change the way they behave.
Bash arithmetic accepts ASCII/string numbers for input, so there are few reasons to actually use the integer attribute.
Also, variable values can't contain ASCII NULL (ie. 8 bit zero), because regular null terminated C strings are used to implement them.
* Ie one or more key + value pairs.
** Reference variables expand to the value of another variable, whose label is assigned to the reference variable
Append a string:
$ foo=Hello
$ foo+=' world!'
$ echo "$foo"
Hello world!
$ num=3
$ num+=4
echo "$num"
34 # Appended string (not a sum)
One of the few reasons to use the integer attribute, is that it changes the behaviour of the += assignment operator:
$ declare -i num=3
$ num+=4
echo "$num"
7 # Sum
Note that this doesn't work for -=, /=, etc. unless you do it inside arithmetic ((( )) and $(( ))), where numbers are already treated the same with or without the integer attribute. See the section "arithmetic evaluation" of man bash for a full list of those operators, which are the same as for C.
The += assignment operator can also be used to append new elements to an indexed array (AKA "list"):
$ foo=(one)
$ foo+=(two)
$ printf 'Separate element: %s\n' "${foo[@]}"
Separate element: one
Separate element: two
Another common way to do this is to use a counter:
$ foo[c++]=one
$ foo[c++]=two
POSIX shells do not use the += assignment operator to append strings, so you have to do it like this:
$ foo=Hello
$ foo="$foo world!"
$ echo "$foo"
Hello world!
This is fine in bash too, so it could be considered a more portable syntax.
dan
In my opinion, the simplest way to concatenate two strings is to write a function that does it for you, then use that function.
function concat ()
{
prefix=$1
suffix=$2
echo "${prefix}${suffix}"
}
foo="Super"
bar="man"
concat $foo $bar # Superman
alien=$(concat $foo $bar)
echo $alien # Superman
Anthony Rutledge
I kind of like making a quick function.
#! /bin/sh -f
function combo() {
echo $@
}
echo $(combo 'foo''bar')
Yet another way to skin a cat. This time with functions :D
CommaToast
Retrieved from : http:www.stackoverflow.com/questions/4181703/how-to-concatenate-string-variables-in-bash
반응형
'etc. > StackOverFlow in English' 카테고리의 다른 글
| How do I loop through or enumerate a JavaScript object? (0) | 2023.06.10 |
|---|---|
| Commit only part of a file in Git (0) | 2023.06.10 |
| How do I convert a String to an int in Java? (0) | 2023.06.08 |
| Is there an "exists" function for jQuery? (0) | 2023.06.08 |
| Why is "using namespace std;" considered bad practice? (0) | 2023.06.08 |