print double quotes in shell programming

Question

I want to print double quotes using echo statement in shell programming.

Example:

echo \"$1,$2,$3,$4\";

prints

Answer 1

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).

Answer 2

You should escape the " to make it visible in the output, you can do this :

echo \""$1"\",\""$2"\",\""$3"\",\""$4"\"

Answer 3

Use printf, no escaping is required:

printf '"%s","%s","%s","%s";\n' $1 $2 $3 $4

and the trailing ; gets printed too!