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've read in the docs for PSReadLineOption that I can amend empty lines below the prompt to separate output from the next input. So I've tried the following.
Set-PSReadLineOption -ExtraPromptLineCount 3
As far I can tell, there's not empty lines appearing and I'm uncertain if I'm doing it wrong, if I'm imagining the result differently than intended or whatever is up with this.
I believe what you're trying to accomplish can be done in a simpler way. At the end of your output, just write a newline "`n" to stdout.
Write-Host "`n"
Sequences such as `n which use the back tick ` which is the PowerShell escape character, and a letter to make an escape sequence. These are called special characters. In the specific case of `n, it represents a newline. In the docs I linked, it lists the escape sequences that you can use within PowerShell, to implement these special characters.
I've been messing arround with Powershell and googling various things as I go along. This one is a little hard to put into words that google woule understand. I can get the indevidual lines of a text file in powershell by indexing:
$textFile = Get-Content "myText.txt"
$textFile[0]
This would output the first line of the text file. But when I put the text file in quotes it will output all lines, even with the index
"$textFile[0]"
How can I still get only get the line I want, while wrapping the variable in quotes? If I try "$textFile"[0] it will just give me the whole file as before. The reason I'm trying to do this is because I'm trying to make that one line of the text file part of a bigger string that I can execute
$remote = "Enter-PSSession -ComputerName`", textFile[0]"
Invoke-Expression $remote
This is my way of illustrating what I'm trying to do.
You can use any of the following methods:
# Sub-expression operator
"Some Text $($textFile[0])"
# String format operator
"My Text {0}" -f $textFile[0]
# Concatenation
("Text"+$textFile[0])
Surrounding double quotes tells PowerShell to expand the string inside. Any variables within will be interpolated. Variables begin with $ and their following names can only have certain characters without requiring a special escape. [ would require an escape and since it isn't escaped, PowerShell interprets the variable name ending with the character just before the [. Therefore $textFile is interpolated, the whole file contents are converted into a string, and [0] is appended to the end of the string.
You can see details of the operators at About_Operators.
See About_Variables for how to create a variable including cases with special characters even if that doesn't directly apply here.
I am working with Powershell. My issue is that my file path (which does not exist on a local computer) has an apostrophe in it. Powershell is seeing this as a single quote, so it is giving me the following error: The string is missing the terminator: '. I thought that I could escape the single quote using a backtick, but that gave me the same error.
The error does not occur when I am doing the first line of code, and I don't even need the backtick for that part. I can even see that the contents of the variable matches up with the file path that I am using. It is only when I am doing the invoke-expression part that it is giving me the error.
I am using https://learn.microsoft.com/en-us/powershell/module/microsoft.powershell.utility/invoke-expression?view=powershell-7, so I don't think the second line of the code is the problem.
My code is listed below:
$code = "\\example\example\John_Doe`'s_Folder\example.ps1"
invoke-expression -command $code
I have also tried wrapping the entire file path in double-quotes and single-quotes, but my program did not like that either. I can't remove the apostrophe as we have over a hundred of systems that are directing to John_Doe's_Folder.
Invoke-Expression should generally be avoided; definitely don't use it to invoke a script or external program.
In your case, simply use &, the call operator to invoke your script via the path stored in variable $code (see this answer for background information), in which case the embedded ' needs no escaping at all:
$code = "\\example\example\John_Doe's_Folder\example.ps1"
& $code
As for what you tried:
"\\example\example\John_Doe`'s_Folder\example.ps1" turns into the following verbatim string content:
\\example\example\John_Doe's_Folder\example.ps1
That is, the ` was removed by PowerShell's parsing of the "..." string literal itself, inside of which ` acts as the escape character; since escape sequence `' has no special meaning, the ` is simply removed.
For the ` to "survive", you need to escape the ` char. itself, which you can do with ``:
"\\example\example\John_Doe``'s_Folder\example.ps1"
I can't quite understand how PowerShell parses commands and need your help.
I read the following explanation by Microsoft's about_parsing documentation:
When processing a command, the PowerShell parser operates in expression mode or in argument mode:
In expression mode, character string values must be contained in quotation marks. Numbers not enclosed in quotation marks are treated as numerical values (rather than as a series of characters).
In argument mode, each value is treated as an expandable string unless it begins with one of the following special characters: dollar sign ($), at sign (#), single quotation mark ('), double quotation mark ("), or an opening parenthesis (().
If preceded by one of these characters, the value is treated as a value expression.
I can understand when parsing a command, PowerShell uses either expression mode or argument mode, but I can't quite understand the following examples.
$a = 2+2
Write-Output $a #4(int), expression mode
Write-Output $a/H #4/H(str), argument mode
I wonder PowerShell expands variable first and then decide which mode when parsing, but is it right?
If so, there's another question about data type.
It seems reasonable for me the former command produces integral, but the latter one doesn't. Why can integer 4 be put next to string /H?
I tried this example and it worked. It seems variables turn into string whatever data type they are when expanded. Is it right?
$b = 100
Add-Content C:\Users\Owner\Desktop\$b\test.txt 'test'
I appreciate for your help.
Edited to clarify the point after got the comment
I've got the comment that the both Write-Output examples are argument mode, so can the examples be interpreted like this?
Write-Output "$a"
Write-Output "$a/H"
I'm terribly sorry for too ambiguous question, but I want to know:
In argument mode, double quotations are omitted?
The Write-Output examples are quoted from microsoft's document I linked and it says the first example produces integral, but is it wrong?