按特定列中的一系列值过滤.csv 文件，而不使用 awk 或 sed

Question

I have a csv file where the data is stored like this, with a space as the delimiting character:

181.221.132.87 2020-03-01T06:22:47.775Z "GET / HTTP/1.1" 200 1 "Mozilla/5.0 (Windows NT 6.1; WOW64; Trident/7.0; rv:11.0) like Gecko"

I have to print all the lines where the 5th column (in this case the column with the value "1") has a value greater than 5. The catch is that I am limited in the unix commands I can use and have been told that I specifically cannot use awk or sed. Anything that cannot be accomplished with the list of commands provided to us must be implemented with custom C programs however, the emphasis is to use custom programs as little as possible.

Unix commands I can use are as follows: cat curl cut echo exec egrep find grep head ls paste printf sort tail tr uniq wc

Sorry if a similar question has been asked before but I cannot find a starting point that doesn't include awk or sed

EDIT:

{ egrep " "[5-9]{1}" " file.csv; egrep " "[0-9]{2}" "file.csv; }

The above command seems to give the correct output, however I feel there is a better solution.

Answer 1

Here's what I came up with!

egrep --color '^(("[^"]*"|[^"]\S*)\s+){4}([1-9][0-9]|[6-9])' file.csv

Explanation

• ^ is the start of the line

• ("[^"]*"|[^"]\S*)\s+) is one cell, it's composed of 2 possibilities:

  • "[^"]*" This is a string cell, surrounded by quotes, and which cannot contain any quotes in its body
  • [^"]\S*)\s+ This is a normal cell, which can contain anything except white spaces ( \s is a white space, \S is a non white space)

• {4} We repeat that 4 times, for the first 4 cells

• ([1-9][0-9]|[6-9]) This is your number, composed of 2 possibilities again:
  • [1-9][0-9] It's more than 10
  • [6-9] It's more than 5

As for the color flag, it... adds color to the command, it makes it easier when you're creating a regex on the go, to have a visual representation of what is matched:

On some system the --color is there by default, so you might not see the difference

Answer 2

Without grep

cat log| while read line
do
  v=`echo $line | cut -d'"' --output-delimiter=' ' -f1,3 | tr -s ' '|cut -f4 -d' ' `
  if [ "$v" -gt 5 ]
  then
    echo $line
  fi                               
done

read the file line by line with while read line

split " with cut odd are fields without " even values inside the "

cut -d'"' --output-delimiter=' ' -f1,3

give 181.221.132.87 2020-03-01T06:22:47.775Z 200 1

remove double spaces with tr

cut -d'"' --output-delimiter=' ' -f1,3 | tr -s ' '

gives

181.221.132.87 2020-03-01T06:22:47.775Z 200 1

get the field at pos 4 with cut

cut -d'"' --output-delimiter=' ' -f1,3 | tr -s ' '|cut -f4 -d' '

gives

1

then check against 5 in pure bash [ "$v" -gt 5 ]