linux - 如何比较unix中不同列的文件？

Question

我想比较 Today.txt 和 Main.txt 的文件名。如果匹配，则打印 Main.txt 匹配文件的所有 6 列，并使用新文件 matched.txt。

以及与Main.txt不匹配的文件，然后在新文件say unmatched.txt中列出TODAY.txt的文件名和时间

主.txt

 date      filename          timestamp space  count   status
Nov 4    +CHCK01_20161104.txt  06:39   2.15M  17153    on_time
Nov 4    TRIPS11_20161104.txt 09:03   0.00M  24       On_Time
Nov 4    AR02_20161104.txt    09:31   0.00M  7        On_Time
Nov 4    AR01_20161104.txt    09:31   0.04M  433      On_Time

今天.txt

 filename       time
CHCK01_20161104.txt 06:03
CHCK05_20161104.txt 11:10
CHCK09_20161104.txt 21:46
AR01_20161104.txt   09:36
AR02_20161104.txt   09:36
ifs01_20161104.txt  21:16
TRIPS11_20161104.txt 09:16

要求的输出:匹配.txt

Nov 4    +CHCK01_20161104.txt  06:39   2.15M  17153    on_time
Nov 4    TRIPS11_20161104.txt 09:03   0.00M  24       On_Time
Nov 4    AR02_20161104.txt    09:31   0.00M  7        On_Time
Nov 4    AR01_20161104.txt    09:31   0.04M  433      On_Time

不匹配.txt

CHCK05_20161104.txt 11:10
CHCK09_20161104.txt 21:46
ifs01_20161104.txt  21:16

你能帮我解决这个问题吗？

提前致谢!

Answer 1

awk 助您一臂之力!

$ awk 'FNR==1{next} 
      NR==FNR{a[$1]=$2; next} 
      $3 in a{print; delete a[$3]} 
          END{for(k in a) print k,a[k] > "unmatched"}' today main > matched

$ head *matched

==> matched <==
Nov 4    CHCK01_20161104.txt  06:39   2.15M  17153    on_time
Nov 4    TRIPS11_20161104.txt 09:03   0.00M  24       On_Time
Nov 4    AR02_20161104.txt    09:31   0.00M  7        On_Time
Nov 4    AR01_20161104.txt    09:31   0.04M  433      On_Time

==> unmatched <==
ifs01_20161104.txt 21:16
CHCK09_20161104.txt 21:46
CHCK05_20161104.txt 11:10