使用awk或sed删除字段以获取值

Question

I have a text file with following data.

            \ {
                Name     "ABC",
                count   378
            }
            \ {
                Name     "DEF",
                count   5283
            }
            \ {
                Name     "BCD",
                count   152244
            }
            \ {
                Name     "XYZ",
                count   5688
            }
            \ {
                Name     "1A2B",
                count   1749132
            }

I want the result like :--

            ABC , 378
            DEF , 5283
            BCD , 152244
            XYZ , 5688
            1A2B ,1749132

I tried to remove the non revelant data using the command :--

            grep -e '^ ' result.txt 

But I am unable to proceed beyond it . Can someone help me with same ?

Answer 1

try following awk and let me know if this helps you.

awk '/Name/{gsub(/\"|\,/,"",$2);val=$2;next} /count/{print val " , " $2}'  Input_file

OR

awk -F'[",]' '/Name/{val=$2;next} /count/{split($0, a," ");print val,a[2]}' OFS=" , "  Input_file

Answer 2

Here's another awk thought you may apply,

$ awk '$2~/[0-9A-Z]/ {printf gsub(/"/,"",$2)?$2:" "$2"\n"}' file

Brief explanation,

• $2~/[0-9A-Z]/ : find the record matched regex [0-9A-Z]
• gsub(/"/,"",$2) : remove " in the $2 , and then print it

Answer 3

If you are using an awk that supports regular expression RS (at least gawk and mawk), you can do it like this:

awk '!(NR%2) { print $3 " , " $5  }' RS='\\ *{|}' FS='[\n," ]+' infile

Output:

ABC , 378
DEF , 5283
BCD , 152244
XYZ , 5688
1A2B , 1749132

Answer 4

sed solution:

sed -En '/[{}]/d;N; s/Name *"([^"]+)".*count *([0-9]+).*/\1 , \2/p;' file

The output:

        ABC , 378
        DEF , 5283
        BCD , 152244
        XYZ , 5688
        1A2B , 1749132