bash：从字符串中删除空格

Question

I've written a small library function to help me exit when the script owner isn't root:

#!/bin/bash   

# Exit for non-root user

exit_if_not_root() {
        if [ "$(id -u)" == "0" ]; then return; fi
        if [ -n "$1" ];
        then
                printf "%s" "$1"
        else
                printf "Only root should execute This script. Exiting now.\n"
        fi
        exit 1
}

Here I call it from another file:

#!/bin/bash

source ../bashgimp.sh

exit_if_not_root "I strike quickly, being moved. But thou art not quickly moved to strike.\n You're not root, exiting with custom message."

And the output is:

I strike quickly, being moved. But thou art not quickly moved to strike.\n You're not root, exiting with custom message.

How can I get the newline to appear correctly?

Answer 1

只需使用echo -e而不是printf

Answer 2

Maybe do away with the "%s" and just

printf "$1"

would be simplest in this case.

ANSI-C escape sequences are not treated as such by default in strings - they are the same as other literals. The form

$'ANSI text here'

will undergo backslash-escape replacement though. ( Ref. )

In your case though, as you're just print ing the formatted string provided, you may as well treat it as the format string rather than an argument.

Answer 3

Or quote the string correctly. You can have literal newlines in a quoted string, or use eg the $'string' quoting syntax.

Answer 4

In the function, change the line to

printf "%s\n" "$@"

And call the function like this

exit_if_not_root "text for line 1" "text for line2" "text for line 3"

printf will re-use the format string for as many value strings as you supply.