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bash - print double quotes in shell programming

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.

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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).


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