Question

我想在4列中打印输出文本：

SharedCacheMap 0xb89e9720   None   \Device\HarddiskVolume1\Windows\System32\WWanAPI.dll
ImageSectionObject 0xb89ea5f8   None   \Device\HarddiskVolume1\Program Files\McAfee\Host Intrusion Prevention\Resource\0409\McTrayHipRL.dll    
DataSectionObject 0xb89ea5f8   None   \Device\HarddiskVolume1\Program Files\McAfee\Host Intrusion Prevention\Resource\0409\McTrayHipRL.dll

我试过了：

column -s " " -t

我不知道如何处理文件路径中的空格。

提前感谢您的帮助！

Answer 1

试试这个：

while read -r c1 c2 c3 rest; do printf "%-20s %-12s %-8s %s\n" "$c1" "$c2" "$c3" "$rest"; done < file

输出：

SharedCacheMap       0xb89e9720   None     \Device\HarddiskVolume1\Windows\System32\WWanAPI.dll
ImageSectionObject   0xb89ea5f8   None     \Device\HarddiskVolume1\Program Files\McAfee\Host Intrusion Prevention\Resource\0409\McTrayHipRL.dll
DataSectionObject    0xb89ea5f8   None     \Device\HarddiskVolume1\Program Files\McAfee\Host Intrusion Prevention\Resource\0409\McTrayHipRL.dll

Answer 2

仅使用sed删除文件名字符串中的空格

cat yourfile.txt | sed 's/[^\]*\\/&\$\n/' | sed '2~2 s/ //g' | sed ':a; N; $! ba; s/\$\n//g'

或用％20替换它们

cat yourfile.txt | sed 's/[^\]*\\/&\$\n/' | sed '2~2 s/ /%20/g' | sed ':a; N; $! ba; s/\$\n//g'

如何在列中打印文本（分隔符=空格，忽略具有空格的特定列）？

2 个答案: