How to Echo a Variable With Single Quotes

How do I use variables in single quoted strings?

Variables are expanded in double quoted strings, but not in single quoted strings:

 $ name=World

 $ echo "Hello $name"
 Hello World

 $ echo 'Hello $name'
 Hello $name

If you can simply switch quotes, do so.

If you prefer sticking with single quotes to avoid the additional escaping, you can instead mix and match quotes in the same argument:

 $ echo 'single quoted. '"Double quoted. "'Single quoted again.'
 single quoted. Double quoted. Single quoted again.

 $ echo '"$name" has the value '"$name"
 "$name" has the value World

Applied to your case:

 echo 'test text "here_is_some_test_text_'"$counter"'" "output"' >> "$FILE"

How can I add single quotes to a variable in PHP

You can do

echo "\"$animal\"";

Or you can just use single quotes instead as the following:

echo '"$animal"';

Whichever you prefer.

How to echo variable inside single quotes using Bash?

JSON string values should be quoted and so should parameter expansions. You can achieve this by using double quotes around the entire JSON string and escaping the inner double quotes, like this:

curl -i -H "Content-Type: application/json" -X POST -d "{\"mountpoint\":\"$final\"}" http://127.0.0.1:5000/connect

As mentioned in the comments, a more robust approach would be to use a tool such as jq to generate the JSON:

json=$(jq -n --arg final "$final" '{ mountpoint: $final }')
curl -i -H "Content-Type: application/json" -X POST -d "$json" http://127.0.0.1:5000/connect

Single quotes around variable value

Like every other command the assignment of a value to a variable will be parsed by the shell, too. To prevent the shell from parsing the special characters must be "quoted". Single characters can be quoted with the backslash (\), multiple characters can be quoted with single quotes (') or double quotes ("). Within single quotes every character will be taken literally. Double-quoted characters will mostly taken literally, with the main exeption (among some others): variables will be replaced with their contents.

To assign the Value "JSBDbshe66!#12$@a" in the command line it must be single-quoted because of the bang (!) sign (see bash bang for further information):

:~$ DEV_ARM_CLIENT_SECRET='JSBDbshe66!#12$@a'

The command

:~$ echo $DEV_ARM_CLIENT_SECRET

gives you the desired output:

JSBDbshe66!#12$@a

To copy this value to another variable just enter in the command line:

ARM_CLIENT_SECRET=$DEV_ARM_CLIENT_SECRET

Alternatively you can enter:

ARM_CLIENT_SECRET="$DEV_ARM_CLIENT_SECRET"

to receive the same result. The variable-name between the double quotes will be replaced with the value and finally the shell removes the Quotes. If you put the variable name between single quotes

ARM_CLIENT_SECRET='$DEV_ARM_CLIENT_SECRET'

every character between the quotes will be taken literally:

:~$ echo $ARM_CLIENT_SECRET

$DEV_ARM_CLIENT_SECRET

If you quote the single quotes via Backslash

:~$ ARM_CLIENT_SECRET=\'$DEV_ARM_CLIENT_SECRET\'

the value of "$ARM_CLIENT_SECRET" will be the same as $DEV_ARM_CLIENT_SECRET with an additional leading and a trailing single quote:

:~$ echo $ARM_CLIENT_SECRET

'JSBDbshe66!#12$@a'

Expansion of variables inside single quotes in a command in Bash

Inside single quotes everything is preserved literally, without exception.

That means you have to close the quotes, insert something, and then re-enter again.

'before'"$variable"'after'
'before'"'"'after'
'before'\''after'

Word concatenation is simply done by juxtaposition. As you can verify, each of the above lines is a single word to the shell. Quotes (single or double quotes, depending on the situation) don't isolate words. They are only used to disable interpretation of various special characters, like whitespace, $, ;... For a good tutorial on quoting see Mark Reed's answer. Also relevant: Which characters need to be escaped in bash?

Do not concatenate strings interpreted by a shell

You should absolutely avoid building shell commands by concatenating variables. This is a bad idea similar to concatenation of SQL fragments (SQL injection!).

Usually it is possible to have placeholders in the command, and to supply the command together with variables so that the callee can receive them from the invocation arguments list.

For example, the following is very unsafe. DON'T DO THIS

script="echo \"Argument 1 is: $myvar\""
/bin/sh -c "$script"

If the contents of $myvar is untrusted, here is an exploit:

myvar='foo"; echo "you were hacked'

Instead of the above invocation, use positional arguments. The following invocation is better -- it's not exploitable:

script='echo "arg 1 is: $1"'
/bin/sh -c "$script" -- "$myvar"

Note the use of single ticks in the assignment to script, which means that it's taken literally, without variable expansion or any other form of interpretation.

How to echo single quotes from within shell in find exec

Assign a quote to a temp variable (I also remove the path).

q="'" find . -iname "*" -exec sh -c 'echo "file $q${0##*/}$q"' {} \;

How to escape single quotes within single quoted strings

If you really want to use single quotes in the outermost layer, remember that you can glue both kinds of quotation. Example:

 alias rxvt='urxvt -fg '"'"'#111111'"'"' -bg '"'"'#111111'"'"
 #                     ^^^^^       ^^^^^     ^^^^^       ^^^^
 #                     12345       12345     12345       1234

Explanation of how '"'"' is interpreted as just ':

1. ' End first quotation which uses single quotes.
2. " Start second quotation, using double-quotes.
3. ' Quoted character.
4. " End second quotation, using double-quotes.
5. ' Start third quotation, using single quotes.

If you do not place any whitespaces between (1) and (2), or between (4) and (5), the shell will interpret that string as a one long word.

Can single quotes be preserved when passing a variable to a command on stdin?

You're doubling up on the parsing, so you need to double up on the quoting.

$: foo="text with \'single quotes\'"
$: xargs echo <<< "$foo"
text with 'single quotes'

so by the same token,

$: foo="text with 'single quotes'"
$: xargs echo <<< \"$foo\"
text with 'single quotes'

This is dependent on the number of parsing passes, though, which is what you need to understand and accomodate. Applying a two-layer solution to a command with only one layer of parsing has the opposite effect -

$: echo \"$foo\"
"text with 'single quotes'"

Now the extra quotes become visible.

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