awk，^反斜杠不是最后一个字符在线

Question

I am trying to use gawk in windows to parse a directory which contains CSV files for the blank rows in the second column in each file. I want to take the whole row where the the 2nd column is blank from ALL the source files and output to a csv. .. what I have so far is below.

I am sure this has to do with properly escaping the single quotes, I've tried using ^ and \\ to no avail. Thanks for any tips.

awk -F, \'!length$2\' *.csv > output.csv 

Answer 1

Thanks to Etan Reisner and HuStmpHrrr I have it working now. If you folks want to post answers I can mark them so that you get reputation points. Your hints pointed me in the right direction. I now have all the rows with a blank 2nd row in one file. The final working version of the command in my case is :

awk -F, !length($2) *.csv > output.csv    //Works ( in windows)

• I am not sure why my answer was edited, but the string above doen`t work when you enclose it in single quotes.

  awk -F, '!length($2)' *.csv > output.csv //Doesn`t Work (in windows)