Extract values from specific row, column using bash

问题

I have a sample log file with fields and values in this format

value1 value2
 50    100

value3 value4 
 10      150   

value5
 200

I need to extract fields and values to something of this format

value1=50
value2=100
value3=10
value4=150
value5=200

回答1:

Try this awk:

gawk '{split($0,n_arr," "); getline; n=split($0,v_arr," "); getline; for (i=1;i<=n;i++){print n_arr[i] "=" v_arr[i]}}'

回答2:

awk '/value/ {
                 if ((getline values) > 0) {
                     split(values, array)
                     for (i = 1; i <= NF; i++)
                         print $i "=" array[i]
                 }
              }' inputfile

Instead of /value/ you might use /[[:alpha:]]/

回答3:

awk '
    NF==0 {next}
    var[1] == "" {for (i=1; i<=NF; i++) var[i]=$i; next}
    {for (i=1; i<=NF; i++) {print var[i] "=" $i; delete var[i]}}
'

回答4:

This might work for you:

sed '/^$/d;N;:a;s/\([^ \n]*\) *\([^\n]* *\n\) *\([^ \n]*\) *\(.*\)/\2\4\n\1=\3/;/^\n\n/!ba;s///' file

来源：https://stackoverflow.com/questions/10501030/extract-values-from-specific-row-column-using-bash