How can I escape a double quote inside double quotes?

Question

How can I escape double quotes inside a double string in Bash?

For example, in my shell script

#!/bin/bash

dbload=\"load data local infile \\\"\'gfp         


        

Answer 1

A simple example of escaping quotes in the shell:

$ echo 'abc'\''abc'
abc'abc
$ echo "abc"\""abc"
abc"abc

It's done by finishing an already-opened one ('), placing the escaped one (\'), and then opening another one (').

Alternatively:

$ echo 'abc'"'"'abc'
abc'abc
$ echo "abc"'"'"abc"
abc"abc

It's done by finishing already opened one ('), placing a quote in another quote ("'"), and then opening another one (').

More examples: Escaping single-quotes within single-quoted strings

Answer 2

Make use of $"string".

In this example, it would be,

dbload=$"load data local infile \"'gfpoint.csv'\" into table $dbtable FIELDS TERMINATED BY ',' ENCLOSED BY '\"' LINES TERMINATED BY \"'\n'\" IGNORE 1 LINES"

Note(from the man page):

A double-quoted string preceded by a dollar sign ($"string") will cause the string to be translated according to the current locale. If the current locale is C or POSIX, the dollar sign is ignored. If the string is translated and replaced, the replacement is double-quoted.