awk打印奇数列

lklkxcxc

echo “a b c d”|awk 怎么构建打印奇数列思路

yinyuemi

echo {1..10} |awk -vOFS= '{for(i=1;i<=NF;i+=2)$(i+1)=FS}1'
1 3 5 7 9

white2010

1. echo 'a b c d e'| awk -v RS=' ' 'NR{i++;if(i%2==1){a[i]=$0}}END{for(j in a)print a[j]}'
   
2. a
   
3. c
   
4. e
   

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winway1988

1. [winway@winway ~]$ echo "a b c d" | awk '{for(i=1;i<=NF;i++) printf("%c%c", i%2?$i:"", i!=NF?" ":"\n")}'
   
2. a  c

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2楼的思路不错

yangkyo821

1. awk '{a=$1;for(i=3;i<=NF;i+=2)a=a FS $i;print a}'

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alan9101

echo 'a b c d e f g' | awk -F' ' '{for(i=1;i<=NF;i+=2)printf("%c\n",$i)}'