bash脚本中的awk过程

Question

I have a task: Write a script for the summation of integers stored in the file. Form a call script: example: sum a.txt 3 4

The input file can contain several columns of integer. The individual columns are separeted be speces ora tabs. The script should sum a appropriate columns and write the result to stdout. So when we have "sum a.txt 3 4" we need to add the number of the third and fourth columns file.

So i do this:

#!/bin/bash
array1=( "$@" )
let LA=${#array1[@]}-1
awk '{for(i=1;i<=$LA;i++)y+=$'${array1[i]}'; print y}' a.txt

but i have an error: awk: : 1unexpected character '.'

Please help is there another way to add up the number of columns whose number are given in the procedure call script?

Answer 1

if you are passing 3 arguments to sum

#!/bin/bash
awk -v col1=$2 -v col2=$3 '{sum1 += $col1; sum2 += $col2} END{print sum1,sum2}' $1

when the script is invoked as sum a.txt 3 4 , then

$1 = a.txt
$2 = 3
$4 = 4

EDIT

If the number of columns can vary

#!/bin/bash

filename=$1
cols=${@:2:$#}

awk -v c="$cols" '{split(c, cols, " "); for (col in  cols) sum[col]+=$col} END{for (i in sum) print sum[i]}' $filename

Explanation

• -vc="$cols" creates an awk variable c and pass the column list

• split(c, cols, " ") split the column list c to obtain an array , cols

• for (col in cols) sum[col]+=$col for each column value in cols array find the sum, sum[col]+=$col

• END{for (i in sum) print sum[i]} print the sum array