strange echo output

问题:

Can anybody explain this behaviour of the bash shell which is driving me nuts

[root@ns1 bin]# export test=`whois -h whois.lacnic.net 187.14.6.108 | grep -i inetnum: | awk '{print $2}'`

[root@ns1 bin]# echo $test
187.12/14

[root@ns1 bin]# echo "iptables -I INPUT -s $test -J DROP"

-J DROP -I INPUT -s 187.12/14

[root@ns1 bin]# 

Why is my echo screwed up? It is being changed by the contents of $test.

If you change $test to "ABC" all is fine. Is it related to the slash?

回答1:

Why is my echo screwed up? It is being changed by the contents of $test.

Because your test contains a carriage return. Remove it:

test=$(whois -h whois.lacnic.net 187.14.6.108 | grep -i inetnum: | awk '{print $2}' | tr -d '\r')

回答2:

Your test contains something like

1234567 -I INPUT -s 187.12/14\r-J DROP

which, due to the carriage return, is visible only as

-J DROP -I INPUT -s 187.12/14

The CR moves the cursor to the start-of-line, where it then overwrites previous characters.

You could try

echo "$test" | od -bc

to verify this.

回答3:

This is almost certainly a carriage return. echo is doing its job correctly and emitting the string to your terminal; the problem is that your terminal is treating a part of the string as a command for it to follow (specifically, a LF character, $'\r', telling it to send the cursor to the beginning of the existing line).

If you want to see the contents of $test in a way which doesn't allow your terminal to interpret escape sequences or other control characters, run the following (note that the %q format string is a bash extension, not available in pure-POSIX systems):

printf '%q\n' "$test"

This will show you the precise contents formatted and escaped for use by the shell, which will be illuminative as to why they are problematic.

To remove $'\r', which is almost certainly the character giving you trouble, you can run the following parameter expansion:

test=${test//$'\r'/}

Unlike solutions requiring piping launching an extra process (such as tr), this happens inside the already-running bash shell, and is thus more efficient.