In Bash, what are the differences between single quotes ('') and double quotes ("")?
Single quotes won't interpolate anything, but double quotes will. For example: variables, backticks, certain \ escapes, etc.
Example:
$ echo "$(echo "upg")"
upg
$ echo '$(echo "upg")'
$(echo "upg")
The Bash manual has this to say:
3.1.2.2 Single Quotes
Enclosing characters in single quotes (') preserves the literal value of each character within the quotes. A single quote may not occur between single quotes, even when preceded by a backslash.
3.1.2.3 Double Quotes
Enclosing characters in double quotes (") preserves the literal value of all characters within the quotes, with the exception of $, `, \, and, when history expansion is enabled, !. The characters $ and ` retain their special meaning within double quotes (see Shell Expansions). The backslash retains its special meaning only when followed by one of the following characters: $, `, ", \, or newline. Within double quotes, backslashes that are followed by one of these characters are removed. Backslashes preceding characters without a special meaning are left unmodified. A double quote may be quoted within double quotes by preceding it with a backslash. If enabled, history expansion will be performed unless an ! appearing in double quotes is escaped using a backslash. The backslash preceding the ! is not removed.
The special parameters * and # have special meaning when in double quotes (see Shell Parameter Expansion).
The accepted answer is great. I am making a table that helps in quick comprehension of the topic. The explanation involves a simple variable a as well as an indexed array arr.
If we set
a=apple # a simple variable
arr=(apple) # an indexed array with a single element
and then echo the expression in the second column, we would get the result / behavior shown in the third column. The fourth column explains the behavior.
#
Expression
Result
Comments
1
"$a"
apple
variables are expanded inside ""
2
'$a'
$a
variables are not expanded inside ''
3
"'$a'"
'apple'
'' has no special meaning inside ""
4
'"$a"'
"$a"
"" is treated literally inside ''
5
'\''
invalid
can not escape a ' within ''; use "'" or $'\'' (ANSI-C quoting)
6
"red$arocks"
red
$arocks does not expand $a; use ${a}rocks to preserve $a
7
"redapple$"
redapple$
$ followed by no variable name evaluates to $
8
'\"'
\"
\ has no special meaning inside ''
9
"\'"
\'
\' is interpreted inside "" but has no significance for '
10
"\""
"
\" is interpreted inside ""
11
"*"
*
glob does not work inside "" or ''
12
"\t\n"
\t\n
\t and \n have no special meaning inside "" or ''; use ANSI-C quoting
13
"`echo hi`"
hi
`` and $() are evaluated inside "" (backquotes are retained in actual output)
14
'`echo hi`'
`echo hi`
`` and $() are not evaluated inside '' (backquotes are retained in actual output)
15
'${arr[0]}'
${arr[0]}
array access not possible inside ''
16
"${arr[0]}"
apple
array access works inside ""
17
$'$a\''
$a'
single quotes can be escaped inside ANSI-C quoting
18
"$'\t'"
$'\t'
ANSI-C quoting is not interpreted inside ""
19
'!cmd'
!cmd
history expansion character '!' is ignored inside ''
20
"!cmd"
cmd args
expands to the most recent command matching "cmd"
21
$'!cmd'
!cmd
history expansion character '!' is ignored inside ANSI-C quotes
See also:
ANSI-C quoting with $'' - GNU Bash Manual
Locale translation with $"" - GNU Bash Manual
A three-point formula for quotes
If you're referring to what happens when you echo something, the single quotes will literally echo what you have between them, while the double quotes will evaluate variables between them and output the value of the variable.
For example, this
#!/bin/sh
MYVAR=sometext
echo "double quotes gives you $MYVAR"
echo 'single quotes gives you $MYVAR'
will give this:
double quotes gives you sometext
single quotes gives you $MYVAR
Others explained it very well, and I just want to give something with simple examples.
Single quotes can be used around text to prevent the shell from interpreting any special characters. Dollar signs, spaces, ampersands, asterisks and other special characters are all ignored when enclosed within single quotes.
echo 'All sorts of things are ignored in single quotes, like $ & * ; |.'
It will give this:
All sorts of things are ignored in single quotes, like $ & * ; |.
The only thing that cannot be put within single quotes is a single quote.
Double quotes act similarly to single quotes, except double quotes still allow the shell to interpret dollar signs, back quotes and backslashes. It is already known that backslashes prevent a single special character from being interpreted. This can be useful within double quotes if a dollar sign needs to be used as text instead of for a variable. It also allows double quotes to be escaped so they are not interpreted as the end of a quoted string.
echo "Here's how we can use single ' and double \" quotes within double quotes"
It will give this:
Here's how we can use single ' and double " quotes within double quotes
It may also be noticed that the apostrophe, which would otherwise be interpreted as the beginning of a quoted string, is ignored within double quotes. Variables, however, are interpreted and substituted with their values within double quotes.
echo "The current Oracle SID is $ORACLE_SID"
It will give this:
The current Oracle SID is test
Back quotes are wholly unlike single or double quotes. Instead of being used to prevent the interpretation of special characters, back quotes actually force the execution of the commands they enclose. After the enclosed commands are executed, their output is substituted in place of the back quotes in the original line. This will be clearer with an example.
today=`date '+%A, %B %d, %Y'`
echo $today
It will give this:
Monday, September 28, 2015
Since this is the de facto answer when dealing with quotes in Bash, I'll add upon one more point missed in the answers above, when dealing with the arithmetic operators in the shell.
The Bash shell supports two ways to do arithmetic operation, one defined by the built-in let command and the other the $((..)) operator. The former evaluates an arithmetic expression while the latter is more of a compound statement.
It is important to understand that the arithmetic expression used with let undergoes word-splitting, pathname expansion just like any other shell commands. So proper quoting and escaping need to be done.
See this example when using let:
let 'foo = 2 + 1'
echo $foo
3
Using single quotes here is absolutely fine here, as there isn't any need for variable expansions here. Consider a case of
bar=1
let 'foo = $bar + 1'
It would fail miserably, as the $bar under single quotes would not expand and needs to be double-quoted as
let 'foo = '"$bar"' + 1'
This should be one of the reasons, the $((..)) should always be considered over using let. Because inside it, the contents aren't subject to word-splitting. The previous example using let can be simply written as
(( bar=1, foo = bar + 1 ))
Always remember to use $((..)) without single quotes
Though the $((..)) can be used with double quotes, there isn't any purpose to it as the result of it cannot contain content that would need the double quote. Just ensure it is not single quoted.
printf '%d\n' '$((1+1))'
-bash: printf: $((1+1)): invalid number
printf '%d\n' $((1+1))
2
printf '%d\n' "$((1+1))"
2
Maybe in some special cases of using the $((..)) operator inside a single quoted string, you need to interpolate quotes in a way that the operator either is left unquoted or under double quotes. E.g., consider a case, when you are tying to use the operator inside a curl statement to pass a counter every time a request is made, do
curl http://myurl.com --data-binary '{"requestCounter":'"$((reqcnt++))"'}'
Notice the use of nested double quotes inside, without which the literal string $((reqcnt++)) is passed to the requestCounter field.
There is a clear distinction between the usage of ' ' and " ".
When ' ' is used around anything, there is no "transformation or translation" done. It is printed as it is.
With " ", whatever it surrounds, is "translated or transformed" into its value.
By translation/ transformation I mean the following:
Anything within the single quotes will not be "translated" to their values. They will be taken as they are inside quotes. Example: a=23, then echo '$a' will produce $a on standard output. Whereas echo "$a" will produce 23 on standard output.
A minimal answer is needed for people to get going without spending a lot of time as I had to.
The following is, surprisingly (to those looking for an answer), a complete command:
$ echo '\'
whose output is:
\
Backslashes, surprisingly to even long-time users of bash, do not have any meaning inside single quotes. Nor does anything else.
I have a DOS batch file that has a line that executes a powershell script. First I tried a very simple script with this line in the batch file:
powershell -command "get-date" < nul
That worked great. But the script has nested double-quote characters, which can sometimes be escaped with a backtick (`) character. So then I tried this:
powershell -command "Write-Host `"hello world`"" < nul
That also worked great. However, the script I need to run is pretty complicated and has more than one level of nested double-quote characters. I have taken the complicated script and simplified it to an example that has the same principles here:
[string]$Source = " `"hello world`" ";
Write-Host $Source;
If I save this script inside a PS script file and run it, it works fine, printing out “hello world” including the double quotes, but I need to embed it in the line in the batch file. So I take the script and put it all on one line, and try to insert it into the batch file line, but it doesn’t work. I try to escape the double-quotes, but it still doesn’t work, like this:
powershell -command "[string]$Source = `" `"hello world`" `";Write-Host $Source;" < nul
Is there a way to do what I want? You might ask why I am doing this, but it’s a long story, so I won’t go into the details.
thanks
You'll have to use a combination of batch's escape character and PowerShell's escape character.
In batch, when escaping quotes, you use the common shell backslash (\) to escape those quotes. In Powershell, you use the backtick `.
So if you wanted to use batch to print out a quoted string with Powershell, you need to first batch escape the quote to declare the variable in Powershell, then to ensure the string is quoted you need batch and Powershell escape another quote, and then your add your desired string, ensuring you batch escape first.
For your example, this will work:
powershell -command "[string]$Source = \"`\"hello world`\"\"; Write-Host $Source;"
Here's a break down of the declaration of the $Source variable:
"[string]$Source = # open quote to begin -command parameter declaration
\" # batch escape to begin the string portion
`\" # Powershell+Batch escape
hello world # Your content
`\" # Posh+Batch again
\"; # Close out the batch and continue
more commands " # Close quote on -command parameter
This renders the string like this in batch:
`"hello world`"
One note, you don't need to explicitly cast $Source as a string since you are building it as a literal string from scratch.
$Source = "string stuff" will work as intended.
I'm having trouble adding a new line into a string in powershell:
Get-ChildItem "C:\Users\matt\Desktop\CShell Install" |foreach {'<Component Id="'+$_.name+'" Guid="' +[guid]::NewGuid() + '">`r`n<File Id="'+$_.name+'" Source="$(var.CShell.TargetPath)"></File></Component>'}
As you can see I want a newline to occur at
``r`n
instead they are printed literally.
Any pointers?
Do not use Single Quote, where you want PowerShell to honor backticks (or any other characters that you need PowerShell to interpret).
"FirstLine`r`nSecondLine" prints
FirstLine
SecondLine
'"FirstLine`r`nSecondLine"' prints
"FirstLine`r`nSecondLine"
I have a replace statement in my code whereby Band's is being replaced by Cig's. However when I put single quote it took the first sentence... Example 'Band'
I tried to use double quote but it does not work. Do you know how to escape the single quote sign?
-replace 'Band's', 'Cig's'
See Escape characters, Delimiters and Quotes and Get-Help about_Quoting_Rules from the built-in help (as pointed out by as Nacimota).
To include a ' inside a single-quoted string, simply double it up as ''. (Single-quote literals don't support any of the other escape characters.)
> "Band's Toothpaste" -replace 'Band''s', 'Cig''s'
Or, simply switch to double-quotes. (Double-quote literals are required when wishing to use interpolation or escape characters.)
> "Band's Toothpaste" -replace "Band's", "Cig's"
(Don't forget that -replace uses a regular expression)
Escape a single quote with two single quotes:
"Band's Toothpaste" -replace 'Band''s', 'Cig''s'
Also, this is a duplicate of
Can I use a single quote in a Powershell 'string'?
For trivial cases, you can use embedded escape characters. For more complex cases, you can use here-strings.
$Find = [regex]::escape(#'
Band's
'#)
$Replace = #'
Cig's
'#
"Band's Toothpaste" -replace $Find,$Replace
Then put the literal text you want to search for and replace in the here-strings.
Normal quoting rules don't apply within the here-string #' - '# delimiters, so you can put whatever kind of quotes you want, wherever you want them without needing any escape characters.
The [regex]::excape() on $Find will take care of doing the backslash escapes on any regex reserved characters that might be in the search pattern.
I want to include an apostrophe in my string. Is it possible to do without using double quotes?
'This is a quote. Can`'t I just include a single quote in it?'
'This is another quote that doesn\'t work'
'Escape a single quote '' using a double single quote'
See the help for the quoting rules.
You can check out the help in the powershell command line by typing:
Get-Help about_Quoting_Rules
It explains that backticks are interpreted literally in single-quoted strings.
Because the contents of single-quoted strings are interpreted literally, you cannot use the backtick character to force a literal character interpretation in a single-quoted string.