Question

我的目录组织如下：

1.0ps_DCA_md_center.xtc
1.0ps_DCA_npt.gro
1.2ps_DCA_md_center.xtc
1.2ps_DCA_npt.gro
1.8ps_DCA_md_center.xtc
1.8ps_DCA_npt.gro
2.0ps_DCA_md_center.xtc
2.0ps_DCA_npt.gro

我想用bash输出：

1.0ps_DCA_npt.gro
1.0ps_DCA_md_center.xtc
1.2ps_DCA_npt.gro
1.2ps_DCA_md_center.xtc
1.8ps_DCA_npt.gro
1.8ps_DCA_md_center.xtc
2.0ps_DCA_npt.gro
2.0ps_DCA_md_center.xtc

请注意在.gro和.xtc文件之间进行切换。

到目前为止，我已经尝试了for循环的几种组合，但是没有任何效果。如果有人可以帮助我，我将不胜感激。

for i in *.gro; do echo $i; for j in *.xtc; do echo $j; done; done

1.0ps_DCA_npt.gro
1.0ps_DCA_md_center.xtc
1.2ps_DCA_md_center.xtc
1.8ps_DCA_md_center.xtc
2.0ps_DCA_md_center.xtc
1.2ps_DCA_npt.gro
1.0ps_DCA_md_center.xtc
1.2ps_DCA_md_center.xtc
1.8ps_DCA_md_center.xtc
2.0ps_DCA_md_center.xtc
1.8ps_DCA_npt.gro
1.0ps_DCA_md_center.xtc
1.2ps_DCA_md_center.xtc
1.8ps_DCA_md_center.xtc
2.0ps_DCA_md_center.xtc
2.0ps_DCA_npt.gro
1.0ps_DCA_md_center.xtc
1.2ps_DCA_md_center.xtc
1.8ps_DCA_md_center.xtc
2.0ps_DCA_md_center.xtc

Answer 1

您可以使用此循环：

for i in *.gro; do
   echo "$i"
   echo "${i%_*}"*.xtc
done

1.0ps_DCA_npt.gro
1.0ps_DCA_md_center.xtc
1.2ps_DCA_npt.gro
1.2ps_DCA_md_center.xtc
1.8ps_DCA_npt.gro
1.8ps_DCA_md_center.xtc
2.0ps_DCA_npt.gro
2.0ps_DCA_md_center.xtc

"${i%_*}"将剥离以_开头的字符串的右侧

如何输出单个替代与bash嵌套for循环？

1 个答案: