bash shell pass variable and quotes

Question

I've a variable, wich contains all types of characters

myvar="alfanum;001String"

and I need to pass them to a command with a single quotes in this way

java -jar myapp.jar  HERE I NEED TO PUT ->  'alfanum;001String' with single quotes

but if I do

java -jar myapp.jar "'$myvar'"

it doesn't work.

Note I can't use just double quotes:

java -jar myapp.jar "$myvar"

because $myvar is a user's input and myapp.jar need to retrieve it rounded with a single quotes

Thank you

Quite related: [Expansion of variable inside single quotes in a command in bash shell script](http://stackoverflow.com/q/13799789/1983854). So something like `java -jar myapp.jar ''"$var"''` should do — fedorqui, Aug 12 '16 at 08:47
Unfortunatelly no, otherwise I would not opened this question :) — user2548436, Aug 12 '16 at 09:08
If the *argument* needs to contain single quotes, it's the caller's responsibility to provide the correct data, not yours. If you weren't using `myvar`, would the correct call be `java -jar myapp.jar 'alfanum;001String'` or `java -jar myapp.jar "'alfanum;001String'"`? — chepner, Aug 12 '16 at 11:45

score 2 · Answer 1 · answered Aug 12 '16 at 08:45

2

You can mix several quote types together to form a string in singe quotes:

"'""$myvar""'" will expand to 'alfanum;001String'. You're probably need just "$myvar" here though.

answered Aug 12 '16 at 08:45

featuredpeow

score 0 · Accepted Answer · answered Aug 12 '16 at 08:37

0

You can do it like this

java --jar myapp.jar $(echo "'$myvar'")

But why not just use double quotes?

answered Aug 12 '16 at 08:37

sge

1

The command substitution needs to be quoted, but in `bash`, this is exactly the same as `java -jar myapp.jar "'$myvar'"`. – chepner Aug 12 '16 at 11:49

2 Answers2