linux bash - 比较两个文件并删除具有相同结尾的重复行

Question

我有两个包含文件路径的文件。

文件 1

/home/anybody/proj1/hello.h
/home/anybody/proj1/engine.h
/home/anybody/proj1/car.h
/home/anybody/proj1/tree.h
/home/anybody/proj1/sun.h

文件 2

/home/anybody/proj2/module/include/cat.h
/home/anybody/proj2/module/include/engine.h
/home/anybody/proj2/module/include/tree.h
/home/anybody/proj2/module/include/map.h
/home/anybody/proj2/module/include/sun.h

我需要一个命令，可能使用 grep，它会比较两个文件并输出两个文件的组合，但如果文件名重复，请保留文件 2 中的文件。

预期输出:

/home/anybody/proj1/hello.h
/home/anybody/proj1/car.h
/home/anybody/proj2/module/include/cat.h
/home/anybody/proj2/module/include/engine.h
/home/anybody/proj2/module/include/tree.h
/home/anybody/proj2/module/include/map.h
/home/anybody/proj2/module/include/sun.h

这样我就可以为我的项目的标签数据库生成一个包含文件的列表，但是有些文件在构建过程中是重复的，我不想在我的数据库中有同一个文件的两个副本。

Answer 1

这个 awk 命令应该可以完成这项工作:

awk -F/ 'NR == FNR{a[$NF]=$0; next} !($NF in a); END{for (i in a) print a[i]}' file2 file1

/home/anybody/proj1/hello.h
/home/anybody/proj1/car.h
/home/anybody/proj2/module/include/map.h
/home/anybody/proj2/module/include/cat.h
/home/anybody/proj2/module/include/engine.h
/home/anybody/proj2/module/include/tree.h
/home/anybody/proj2/module/include/sun.h