路径作为循环变量时如何只截取文件名称

 

001、

(base) [root@PC1 test2]# ls
a.txt  b.txt
(base) [root@PC1 test2]# cat a.txt   ## 路径循环变量
./j004_bam/KO_mt_coverage.txt
./aaa_bam/CCP5_mt_coverage.txt
./zzz_bam/DAG7_mt_coverage.txt
./j004_bam/CN_mt_coverage.txt
./j003_bam/ZOM9_mt_coverage.txt
(base) [root@PC1 test2]# cat b.txt
hello world
(base) [root@PC1 test2]# for i in $(cat a.txt ); do name=$(basename $i|sed "s/_mt_coverage.txt//"); awk -v a=$name '{print a,$0}' b.txt; done
KO hello world
CCP5 hello world
DAG7 hello world
CN hello world
ZOM9 hello world

。