如何仅截断给定的列长度？

Question

输入：

a@notebook:~$ cat in.csv
'XYZ843141'^'ASDFSAFXYVFSHGDSDg sdGDS  dsGDSgfa assfd faSDFAS saDFSAFD adFSA343fa sdfSADF'^'BAAAR'^'YYY'^'..... and so on, further columns
'YYZ814384'^'ASfdEtRiuognfnseaFREQTzKb   aSFfdsaADSFSA  adsFdsa34 34 ASFfsas  saftrzj etrzrasdfasffasf safs'^'foooobaaar'^'ZZZ'^'..... and so on, further columns

输出：

a@notebook:~$ cat in.csv | SOMEMAGIC
'XYZ843141'^'ASDFSAFXYVFSHGDSDg s'^'BAAAR'^'YYY'^'..... and so on, further columns
'YYZ814384'^'ASfdEtRiuognfnseaFRE'^'foooobaaar'^'ZZZ'^'..... and so on, further columns

我的问题：如果：

'^'

是分隔符，那么 SOMEMAGIC (awk/sed??) 如何将第二列截断为给定长度？示例：最多 20 个字符，来自：

ASDFSAFXYVFSHGDSDg sdGDS  dsGDSgfa assfd faSDFAS saDFSAFD adFSA343fa sdfSADF

对此：

ASDFSAFXYVFSHGDSDg s

并保留所有其他内容：\

Answer 1

> awk -v OFS="'^'" -F"'\\\\^'" '{if(length($2)>20) $2=substr($2,1,20); print;}' file
'XYZ843141'^'ASDFSAFXYVFSHGDSDg s'^'BAAAR'^'YYY'^'..... and so on, further columns
'YYZ814384'^'ASfdEtRiuognfnseaFRE'^'foooobaaar'^'ZZZ'^'..... and so on, further columns