This question already has an answer here:
-bash: !": event not found [closed]
(1 answer)
Closed 6 years ago.
Simply run
echo "This is a **test** see "
You get:
bash: 
use single quotes
echo 'This is a **test** see '
single quotes take the test you pass as verbatim. While double quoted strings are interpreted by bash, for instance $VAR would indicate a variable, $(..) would execute a command and insert the stdout, and [...] are treated as boolean evaluator and ! as NOT, and so on. From the bash manual:
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 withindouble quotes. The backslash retains its special meaning only when followed by one of the following characters: $, `, ", \, or . 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 ! is interpreted as the NOT (logical NOT) operator and thus [alt is treated as a command. Using single quotes is probably the best option here:
echo 'This is a **test** see '
Related
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.
This question already has answers here:
Difference between single and double quotes in Bash
(7 answers)
Closed 1 year ago.
USER_UID=$1
echo 'generate_token("$USER_UID")'
I want output like
generate_token("1234567")
i tried multiple ways but didn't worked. it just print same line without value generate_jwt("$USER_UID")
When you use single quotes, it causes the shell to preserve the literal value of each character within the quotes. This means the $ will be treated as a literal $ character.
You should use double quotes:
USER_UID="$1"
echo "generate_token(\"$USER_UID\")"
From the bash man page, under the Quoting section:
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, !.
For POSIX details on quoting, see here.
Example in an interactive shell:
$ USER_UID='foo'
$ echo "generate_token(\"$USER_UID\")"
generate_token("foo")
This will also work if USER_UID contains spaces:
$ USER_UID='var with spaces'
$ echo "generate_token(\"$USER_UID\")"
generate_token("var with spaces")
The problem:
By typing, echo "Hello World!", into the terminal I'm am answered with: dquote>
The Question (2-parts):
If I use single quotes, terminal returns the result that I'm expecting to see: Hello World!
How do you escape an !?
Why are double quotes different than single quotes here?
I have tried:
Using backslash, echo "Hello World/!" doesn't seem to work, I'm still returned dquote>.
I've also tried looking for other answers on escape characters and stackoverflow answers. They usually end up going over my head.
About question 1: (How do you escape an !?)
You can use "'!'"
Example:
echo "Hello World"'!'""'!'" I'm there."
Result:
Hello World!! I'm there.
About question 2: (Why are double quotes different than single quotes here?)
Because ! is the default "history expansion" character and inside double quotes "history expansion" is performed.
Bash manual:
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).
In a directory which is added to my $PATH I have the following file:
filename="$1" day="$2" month="$3"
ruby -r "./SomeClass.rb" -e 'SomeClass.run($filename, $day, $month)'
Suppose this file is called someclass. When I type someclass into terminal my system recognizes it as a valid command and runs the corresponding Ruby file correctly. But the arguments are not being passed in. How do I pass arguments into an alias?
How do I pass arguments into an alias?
Don't use an alias, better to declare a function for this. You can pass arguments and do other stuff easily inside a bash function:
someclass() {
# make sure enough arguments are passed in
# echo "$1 - $2 - $3";
filename="$1"; day="$2"; month="$3";
ruby -r "./SomeClass.rb" -e "SomeClass.run($filename, $day, $month)";
}
You are using single quote. Change them to double:
ruby -r "./SomeClass.rb" -e "SomeClass.run($filename, $day, $month)"
No expansion are done within single quotes. From bash manual:
3.1.2.2 Single Quotes
Enclosing characters in single quotes (‘'’) preserves the literal value of each character within the quotes. A single quote may not occur between single quotes, even when preceded by a backslash.
3.1.2.3 Double Quotes
Enclosing characters in double quotes (‘"’) preserves the literal value of all characters within the quotes, with the exception of ‘$’, ‘`’, ‘\’, 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.
A useful page on quotes I am using is on grymoire.
Regarding the title of the question, check this question (Alias with variable in bash) and its answer: you can't and have to use functions.
I want to print double quotes using echo statement in shell programming.
Example:
echo "$1,$2,$3,$4";
prints xyz,123,abc,pqrs
How to print "xyz","123","abc","pqrs";
I had tried to place double quotes in echo statement but its not being printed.
You just have to quote them:
echo "\"$1\",\"$2\",\"$3\",\"$4\""
As noted here:
Enclosing characters in double quotes (‘"’) preserves the literal
value of all characters within the quotes, with the exception of ‘$’,
‘`’, ‘\’, 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).
Use printf, no escaping is required:
printf '"%s","%s","%s","%s";\n' "$1" "$2" "$3" "$4"
and the trailing ; gets printed too!
You should escape the " to make it visible in the output, you can do this :
echo \""$1"\",\""$2"\",\""$3"\",\""$4"\"