Awk: extract content within bcracket

What's a good way to extract foo.开发者_运维百科com from the string using awk?

ImaString (foo1.com, ,bar.com) (foo2.com, , bar.com)

Output i want: foo1.com foo2.com

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If you have more than one field in the comma separated lists:

awk -F, -v RS="[)(]" 'NF>1{print $1}' inputfile

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Apologies for the snarky comment, but the question is not clearly worded. You might try setting FS to the regex "(|," and let awk extract the field for you automatically:

$ awk -F ',|\\(' ...

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Question is still a bit unclear, are you asking for this?

$ echo "ImaString (foo.com, ,bar.com) (foo.com, , bar.com)" | 
  awk -F '(' '{sub(/,.*/,"", $2);{print $2}}'
foo.com

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Are you sure you want to use awk for this? Sed and grep would seem to be a better choice for this.

$ echo "ImaString (foo1.com, ,bar.com) (foo2.com, , bar.com)" | 
  grep -E -o '\(([a-zA-Z0-9]*\....),' | sed -e 's/^.//' -e 's/,$//'
foo1.com
foo2.com

Dump the results into a bash variable, count them with wc, or stick them in a bash array.