unix中的shell脚本错误

Question

what does using 1# for a directory do here?

cpfile=${1#/usr/newconfig} ls -l $cpfile | read var1 echo var1

I am tryiing to understand the script, but I cannot find any resource that would help me ascertain the meaning of this command.

Answer 1

In general, variables can be modified with # , which removes a leading string. If a is a variable with content foobar , then ${a#foo} strips the foo and evaluates to bar . In your case, $1 is a positional parameter, and ${1#/usr/newconfig} is the value of that string with "/usr/newconfig" removed from the start. If it does not match, the expression evaluates to the same string as $1.

However, this script has a rather significant issue in that it assigns cpfile in the environment of the ls that is run, but ls does not receive it as an argument. As a result, ls is going to operate on the current working directory rather than the directory passed as an argument to the script. (unless cpfile is assigned previously in the script somewhere.)