I have a project where I'm trying to dynamically prepare a Powershell script, then write it out to a file for later execution.
Here's a minimal example of the issue:
# The string that I want to write out, having most special characters
$_output_str = "~##$%&*()-_=+{}[]<>?/.,'`"``;!"
# Write it out directly
Write-Output "$_output_str"
This is putting a string with escaped special characters into a variable, then writing the value of that variable out to stdout. It works as expected:
~##$%&*()-_=+{}[]<>?/.,'"`;!
Now, let's try storing that command in a file for later execution.
# The string that I want to write out, having most special characters
$_output_str = "~##$%&*()-_=+{}[]<>?/.,'`"``;!"
# Turn it into a string command
$_cmd = "Write-Output `"$_output_str`""
# Write the command out to a file
"$_cmd" | Out-File "execution.ps1"
And now let's try executing that file:
powershell.exe .\execution.ps1
This throws an error:
At C:\{MYPATH}\execution.ps1:1 char:43
+ Write-Output "~##$%&*()-_=+{}[]<>?/.,'"`;!"
+ ~
The string is missing the terminator: ".
+ CategoryInfo : ParserError: (:) [], ParseException
+ FullyQualifiedErrorId : TerminatorExpectedAtEndOfString
OK, so let's look at the content that was written to the file. This is what execution.ps1 contains:
Write-Output "~##$%&*()-_=+{}[]<>?/.,'"`;!"
Notice that the " and the backtick near the end of the string are no longer escaped. So for some reason, when creating the $_cmd string, it got rid of my escape sequences.
How can I prevent this from happening?
Use single-quoted here-string for the value so it doesn't get expanded.
$_output_str = #'
~##$%&*()-_=+{}[]<>?/.,'"`;!
'#
You can also use regular single-quoted strings but keep in mind that you'll have to double any single-quote part of the string.
# The single quote is escaped by doubling it, meaning the final
# representation will be slightly different
$_output_str = '~##$%&*()-_=+{}[]<>?/.,''"`;!'
# Actual value (the two '' were escaped into a single ')
# '~##$%&*()-_=+{}[]<>?/.,'"`;!'
It is still better than double-quoted strings in any cases.
When you use a double-quoted strings, everything that can get evaluated get evaluated. Your backticks serves to escape the character right after but they are not part of the final representation of the string.
Double-quoted strings
A string enclosed in double quotation marks is an expandable string.
Variable names preceded by a dollar sign ($) are replaced with the
variable's value before the string is passed to the command for
processing.
...
expressions are evaluated, and the result is inserted in the string.
I recommend you take a look at the official documentation for more information. About_Quoting_Rules
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.
The following match returns false. How can I change the regular expression to correct it?
"hello$world" -match '^hello$(wo|ab).*$'
"hello$abcde" -match '^hello$(wo|ab).*$'
'hello$world' -match '^hello\$(wo|ab).*$'
'hello$abcde' -match '^hello\$(wo|ab).*$'
You need to quote the left hand side with single quotes so $world isn't treated like variable interpolation. You need to escape the $ in the right hand side so it isn't treated as end of line.
From About Quoting Rules:
When you enclose a string in double quotation marks (a double-quoted string), variable names that are preceded by a dollar sign ($) are replaced with the variable's value before the string is passed to the command for processing.
...
When you enclose a string in single-quotation marks (a single-quoted string), the string is passed to the command exactly as you type it. No substitution is performed.
From About Regular Expressions:
The two commonly used anchors are ^ and $. The carat ^ matches the start of a string, and $, which matches the end of a string. This allows you to match your text at a specific position while also discarding unwanted characters.
...
Escaping characters
The backslash \ is used to escape characters so they are not parsed by the regular expression engine.
The following characters are reserved: []().\^$|?*+{}.
You will need to escape these characters in your patterns to match them in your input strings.
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?
In my code, I have a certain register which I called "foo" using the m4 macros:
define(foo, w19)
Later on, I'm trying to print a string with the following format:
print: .string "The value of foo is %d\n"
Of course, my output is:
"The value of w19 is 5"
When I want my output to be:
"The value of foo is 5"
How do I escape the macro within this string?
According to the first reference I found, and also the Wikipedia page on m4, expansion is suppressed by quoting, i.e.
`foo' is foo
Note that the opening and closing quote delimiters are backtick and apostrophe respectively.