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In awk how can I replace all double quotes by escaped double quotes?
The dog is "very" beautiful
would become
The dog is \"very\" beautiful
I've seen this answer (Using gsub to replace a double quote with two double quotes?) and I've tried to adapt it, but I'm not very good with awk (and sed is no option because I work both on Linux and OS X and they have different 'sed' installed)
805
One use of an escape sequence is to include a double-quote character in a string constant. Because a plain double quote ends the string, you must use ' \" ' to represent an actual double-quote character as a part of the string. For example: $ awk 'BEGIN { print "He said \"hi!\
If you use single quotes to create a string, you can not use single quotes within that string without escaping them using a backslash ( \ ). The same theory applies to double quotes, and you have to use a backslash to escape any double quotes inside double quotes.
String literalsDouble quote characters (") are escaped by a backslash (\)." Enclose the string in single-quotes ( ' ): 'Another string literal. Single quote characters (') require escaping by a backslash (\).
' 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).
From the answer you linked:
With gsub:
echo 'The dog is "very" beautiful' | gawk '{ gsub(/"/,"\\\"") } 1'
Alternative, with sed:
echo 'The dog is "very" beautiful' | sed 's/"/\\"/g'
In both cases output is:
The dog is \"very\" beautiful
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