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.
Question (because I can't work it out), should ""hello world"" be a valid field value in a CSV file according to the specification?
i.e should:
1,""hello world"",9.5
be a valid CSV record?
(If so, then the Perl CSV-XS parser I'm using is mildly broken, but if not, then $line =~ s/\342\200\234/""/g; is a really bad idea ;) )
The weird thing is is that this code has been running without issue for years, but we've only just hit a record that started with both a left double quote and contained no comma (the above is from a CSV pre-parser).
The canonical format definition of CSV is https://www.rfc-editor.org/rfc/rfc4180.txt. It says:
Each field may or may not be enclosed in double quotes (however
some programs, such as Microsoft Excel, do not use double quotes
at all). If fields are not enclosed with double quotes, then
double quotes may not appear inside the fields. For example:
"aaa","bbb","ccc" CRLF
zzz,yyy,xxx
Fields containing line breaks (CRLF), double quotes, and commas
should be enclosed in double-quotes. For example:
"aaa","b CRLF
bb","ccc" CRLF
zzz,yyy,xxx
If double-quotes are used to enclose fields, then a double-quote
appearing inside a field must be escaped by preceding it with
another double quote. For example:
"aaa","b""bb","ccc"
Last rule means your line should have been:
1,"""hello world""",9.5
But not all parsers/generators follow this standard perfectly, so you might need for interoperability reasons to relax some rules. It all depends on how much you control the CSV format writing and CSV format parsing parts.
That depends on the escape character you use. If your escape character is '"' (double quote) then your line should look like
1,"""hello world""",9.5
If your escape character is '\' (backslash) then your line should look like
1,"\"hello world\"",9.5
Check your parser/environment defaults or explicitly configure your parser with the escape character you need e.g. to use backslash do:
my $csv = Text::CSV_XS->new ({ quote_char => '"', escape_char => "\\" });
I'm trying to insert some data into my table using the copy command :
copy otype_cstore from '/tmp/otype_fdw.csv' delimiter ';' quote '"' csv;
And I have this answer :
ERROR: unterminated CSV quoted field
There is the line in my CSV file where I have the problem :
533696;PoG;-251658240;from id GSW C";
This is the only line a double quote and I can't remove it, so do you have some advice for me ?
Thank you in advance
If you have lines like this in your csv:
533696;PoG;-251658240;from id GSW C";
this actually means/shows the fields are not quoted, which is still perfectly valid csv as long as there are no separators inside the fields.
In this case the parser should be told the fields are not quoted.
So, instead of using quote '"' (which is actually telling the parser the fields are quoted and why you get the error), you should use something like quote 'none', or leave the quote parameter out (I don't know Postgres, so I can't give you the exact option to do this).
Ok, I did a quick lookup of the parameters. It looks like there is not really an option to turn quoting off. The only option left would be to provide a quote character that is never used in the data.
quote E'\b' (backspace) seems to work ok.
Bit late to reply, but I was facing same issue and found that for double quote we need to add one more double quote to escape with along with prefix and suffix double quote to that special characater. for your case it would be
input data is : from id GSW C"
change data to : from id GSW C""""
note there are 4 consecutive double quotes.
first and last double quote is prefix and postfix as per documentation.
middle 2 double quotes is data with one escape double quote.
Hope this helps for readers with similar issue going forward.
So for every double quote in data it needs to be escaped with one escape character (default double quote). This is as per documentation.
I'm using Powershell with ODBC to transfer table data between Sage 50 and MariaDb.
Some data rows in a text column in Sage can contain both single and double quotes that need to be retained when the data is imported into MariaDb.
I'm struggling to get PowerShell to replace for the Values portion of the Insert statement:
I need to replace a single quote ' with backslash single quote \', and also a double quote " with backslash double quote \"
For anyone else searching for this the answer is :
For single quotes
[regex]::replace($text, "'", "\'")
For double quotes
[regex]::replace($DETAILS, "`"", "\'")
I found this URL helpful https://vwiki.co.uk/MySQL_and_PowerShell and as stated you may wish to put this in a function.
How does one escape characters in string literals in Avaloq scripts? I have been unable to find a definitive answer.
I am trying to include the new line character along with a quote in translation in the AMI
Escaping characters is the same as pl/sql, using '' (double single quotes)
And to access the return character, use util.rtn
Like this: '''it''s time''' || util.rtn || 'right?';