Ignoring embeded spaces with AWK

Question

I\'m looking for a simple way to print a specific field with awk while allowing for embedded spaces in the field.

Sample: Field1 Field2 \"Field Three\" Field

Answer 1

Mark Setchell's answer is good, although it will not work if you don't know in advance how many embedded quotes you have (and it doesn't split on spaces anymore).

I hacked this together (obviously it can be improved):

gawk -v FIELD=2 '{ a=$ FIELD; if (substr(a, 0, 1) == "\"") { gsub(/^\"/, "", a); s=a; for (i = FIELD + 1; i <= NF; i++) { a=$ i; nbSub=gsub(/\"$/, "", a); s = s " " a; if (nbSub > 0) { break } } print(s) } }' <<<'allo "hello world" bar'

I would recommend using something else than gawk for this (maybe look into parsing the fields with your shell's IFS variable?).

Addendum: As I said above, this is not really the right tool for the job. For example, you can specify the first field with the -v FIELD=, but it counts fields based on AWK's separator (the embedded spaces are still counted).