如何获得开头处空格的数目？

yufeiluo

字符串   "     xyz  "  , x 前面有5个空格。
如何一句awk ,输出这个字符串，开头到字符出现，有5个空格？

echo  "     xyz  " | awk   ？？？

haooooaaa

1. /tmp$ echo  "     xyz  " | awk -vFPAT='^ +' '{print length($1)}'
   
2. 5
   
3. 
   
4. /tmp$ echo  "     xyz  " | awk 'match($0,/^ +/){print RLENGTH}'
   
5. 5
   

复制代码

jason680

回复 1# yufeiluo

1. don't care
$ echo "     xyz abc" | awk '{print $1}'
xyz

$ echo "     xyz abc" | awk '{print $0}'
     xyz abc

$ echo "     xyz abc" | awk '{$1=$1;print $0}'
xyz abc

2. remove it directly
$ echo "     xyz abc" | awk '{sub(" +","");print $0}'
xyz abc

$ echo "     xyz abc" | awk '{sub(" +","");gsub("x","X");print $0}'
Xyz abc

$ echo "     xyz abc" | awk '{gsub("x","X");print $0}'
     Xyz abc

$ echo "     xyz abc" | awk '{sub(" +","");gsub("x","X");print $0}'
Xyz abc