列出bash参数中的所有目录

Question

假设我有一个bash脚本example.sh。运行该脚本需要一个参数，即目录列表。脚本的工作方式如下：

./example.sh "/dir/location/281/rest-of-the-location"  "/dir/location/287/rest-of-the-location"  "/dir/location/37/rest-of-the-location"  "/dir/location/3007/rest-of-the-location"

所有目录列表都保存在一个名为file.txt的文件中。有没有一种简单的方法可以运行脚本，而不是像上面那样将目录放在第二个引号中？

所有目录都保存在一个名为file.txt的文件中

file.txt
/dir/location/281/rest-of-the-location
/dir/location/37/rest-of-the-location
/dir/location/3007/rest-of-the-location
/dir/location/3127/rest-of-the-location
/dir/location/37/rest-of-the-location
/dir/location/372/rest-of-the-location
/dir/location/137/rest-of-the-location

Answer 1

使用xargs:

xargs ./example.sh < file.txt

如果example.sh与用户交互，您可能需要将-o选项添加到xargs，或者与GNU xargs一起使用xargs -a file.txt ./example.sh