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bashvariablesoutputecho

echo specific string inside quotes from variable

I need a script that echo one of strings wrapped in quotes.

For example, from variable x="C. Ronaldo" "dos Santos Aveiro" "Cristiano Ronaldo" i want to echo the third one.

So I want to get output like: "Cristiano Ronaldo"

I've tried it with echo $x | awk '{print $3}' but it gives me "dos..

Any help?

Solution

  • By default awk splits records by spaces into fields, given your string, "dos is the third field. To parse this string the way you desire GNU awk's FPAT is required. E.g:

    awk -v FPAT='"[^"]*"' '{print $3}'