title: Print columns in csv file starting from last column using Awk
body: <p>I have a CSV file containing thousands of records. Each record is comma delimited with 114 columns each.</p>
  <p>I need to print 4 columns: 89,90,91,92 but column 87 and 86 are single quoted and might have comma or special characters inside which for my knowledge complicates the use of <code>awk</code> to extract such columns.</p>
  <p>I thought it might be simpler to use <code>awk</code> and extract such columns starting from the end of the record: extracting column 23,24,25,26 starting from the last column of the record where no comma inside last column exists.</p>
  <p>Let me know if you can provide any help.</p>
  <p>Example of input record:</p>
  <pre><code>{144361,1,,,,,,,,,,,,'video14.ams01.hls.','http://video14.ams01.hls.ttvnw.net/hls-826494/wolfontwitchtv_16706535008_316775727/mobile/py-index-live.m3u8?token=id=9043516345222409985,bid=16706535008,exp=1443697094,node=video14-1.ams01.hls.justin.tv,nname=video14.ams01,fmt=mobile&amp;sig=af9c82188126e...','AppleCoreMedia/1.0.0.12B436 (iPhone; U; CPU OS 8_1_1 like Mac OS X; en_us)’,column89,column90,column91,column92,’ios','iPhone',298209470,1,,,,,,,,,1,,1,,1,}
  </code></pre>
  <p>Expected result:
  {column89,column90,column91,column92}</p>