Question

这里要求将职位9-12改为空白,例如,在所有固定规模的记录中。

I 旧

cat file | awk  {sub(substr($0,9,12),"   ",$0);print}

工作,除非职位上有一个空白 9. 要求指出这种限制。

cat file | awk -F"*"  {sub(substr($0,9,12),"   ",$0);print}

无帮助。这里的右yn是什么?

Answer 1

Sample file:

$ cat file
1234567890123456
1234567 90123456
1234567890123456
1234567 90123456
1234567890123456

1 D-1, 1 D-1, 1 D-1, 1 D-1, 1 P-5, 1 P-4, 1 P-3, 1 P-2, 1 GS 想法:

awk  { print substr($0,1,8) "    " substr($0,13) }  file

<>0>NOTES:

this replaces each of the 9th-12th positions with a space (ie, replace the 4 fields with a total of 4 spaces)
if the intention is to replace the 4 fields with a single space then replace the " " with " "

这产生了:

12345678    3456
1234567     3456
12345678    3456
1234567     3456
12345678    3456

利用同一想法积极处理这些立场(在评论中类似于Cyrus建议):

awk -v start=9 -v end=12  { print substr($0,1,start-1) sprintf("%*s",(end-start+1)," ") substr($0,end+1) }  file

这也产生了:

12345678    3456
1234567     3456
12345678    3456
1234567     3456
12345678    3456

Answer 2

利用GNU AWK

$ awk -v from=9 -v to=12  BEGIN{FS=OFS=""}{for(i=from;i<=to;i++) $i=" "}1  file