I have the following line in the CSV file

```
   1,LINUX,"/usr/cti/conf/apps/parameters  param1 param2","root passowrd" 

```

after I remove the separators "," by sed

I want now to read each field from the line 

```
#  echo 1  LINUX "/usr/cti/conf/apps/parameters  param1 param2" "root passowrd"  | awk '{print $1}'
1

#  echo 1  LINUX "/usr/cti/conf/apps/parameters  param1 param2" "root passowrd"  | awk '{print $2}'
LINUX

#  echo 1  LINUX "/usr/cti/conf/apps/parameters  param1 param2" "root passowrd"  | awk '{print $3}'
/usr/cti/conf/apps/parameters

```

the expected result on the third field should be **"/usr/cti/conf/apps/parameters param1 param2"** 

Please advice how to read the field in order to get all the words between the Quotation mark , 

what need to add in my awk command ?

So In case of single word then awk should print only this word
In case awk see words between the quotation mark then awk should print all the word between the quotation mark

* remark each field in the CSV could be with Quotation mark or without , 
for example the first field could be with 3 words as “ one two three” or maybe the first field could be only with one word , awk should support both cases