如何按固定长度输出字段

jiaoli

如下文本：
201318984,1.53
201298436,4.69
201226865,9.97
201226873,12.75
201225698,136.98
201365254,1256.36

如何把逗号后的字段固定为八位，不够的用空格补

谢谢！

昭襄王

回复 1# jiaoli

1. sed 's/$/________/' file|cut -c -18
   
2. 201318984,1.53____
   
3. 201298436,4.69____
   
4. 201226865,9.97____
   
5. 201226873,12.75___
   
6. 201225698,136.98__
   
7. 201365254,1256.36_
   

复制代码

yinyuemi

回复 1# jiaoli

1. echo "201318984,1.53
   
2. 201298436,4.69
   
3. 201226865,9.97
   
4. 201226873,12.75
   
5. 201225698,136.98
   
6. 201365254,1256.36" |awk -F, '{printf "%s,%."10-length($2)"f\n",$1,$2}'
   
7. 201318984,1.530000
   
8. 201298436,4.690000
   
9. 201226865,9.970000
   
10. 201226873,12.75000
    
11. 201225698,136.9800
    
12. 201365254,1256.360
    

复制代码

ywlscpl

awk -F, -v OFS=, '{$NF=sprintf("%8s",$NF)}1' file

jiaoli

awk -F, -v OFS=, '{$NF=sprintf("%8s",$NF)}1' file
ywlscpl 发表于 2011-04-25 09:37

   
空格在数字前面了，能不能在数字后面？

blackold

就是格式代啊， printf或 sprintf

yinyuemi

回复 5# jiaoli


    awk -F, '{printf "%s,%-8s\n",$1,$2}'

ywlscpl

回复 5# jiaoli


    %-8s

jiaoli

可以了，谢谢各位！