Shell Programming and Scripting

try using grep

echo adfdffdg | egrep -v "\\$| \*|\&"

u can all unwanted character to this

That's not going to work (especially not without double-quoting the variable). echo does its own parsing, that's why ksh offers print instead. But if you're not on ksh, you don't have that. (The [[ ... ]] construct is also not available in Bourne-compatible syntax, although any POSIX shell should have it.)

Maybe try printf instead, although that, too has some parsing of backslashes and percent signs. This is why I've been suggesting case all along.

In the case statement, you need to double the backslash; *\* means "anything followed by a literal asterisk".

But as already stated, the backslash is actually interpreted by read already. If you have read -r then use that.

PS. Run your spell checker to see how to spell "precede". (^:

It does not seem to work for "\":

>echo "adfdffdg" | egrep -v "\"
egrep: syntax error
> echo "adfdffdg" | egrep -v "/\"
egrep: syntax error

Era, I have to agree that your case statement is the one closest to solving the problem. It's 99% working now.

As for "\", I have incorporated double backslash as follows:

case $b in
*[!-_A-Za-z0-9/]*) echo "Error" >&2;;
/*) echo "Error preceeded by /" >&2;;
*/) echo "Error ended with /" >&2;;
*\\*) echo "Error Also" >&2;;
*) ;;
esac


Unfortunately, the problem remains. See the result below

Enter directory: 12\99 [Still unable to detect the backslash]
Enter directory: 67\\88 [I am not sure why when the input comes with double backslash, the first case statement can detect them]
Error
Enter directory: 562\\\888
Error
Enter directory: 112\67\99

use as follow

echo "asd\\\\asdfsa" | egrep -v "\\\\"

Just discovered something new to me. The value of b, c and INPUT are all the same. Initially I thought I could compare the value of INPUT and b for identifying the backslash. The "\" is automatically removed from the values of b & c. It's a headache now!!!

c=$b
INPUT=`echo $b | cut -d '\' -f1`
echo "INPUT is $INPUT"
echo "b is $b"
echo "c is $c"
if [ "$INPUT" = "$b" ]; then
echo "OK"
else
echo "KO"
fi

Output:

Enter directory: 133\rwtw
INPUT is 133rwtw
b is 133rwtw
c is 133rwtw
OK

See above. The backslash is interpreted and removed already by read but you could see if your shell understands read -r

No offense, but this is the third or forth time I try to point this out.


For understanding the egrep output, you need to understand how the backslash and the single and double quotes are interpreted by the shell and by grep.

Without quoting, a single backslash quotes the next character. echo \ is a syntax error (or rather, escapes the newline after the backslash, and gives you a secondary prompt). echo \\ gives a backslash as the argument to echo, which might echo it, or might complain that a lone backslash is a syntax error. (I get the former case.) echo "\" is a syntax error (or rather, the shell waits for the closing double quote; echo "\"" passes a single double quote, inside double quotes; it needs to be escaped so the shell doesn't treat it as the closing double quote) and echo "\\" again passes a single backslash as the argument to echo.

egrep uses backslashes for escaping (like some versions of echo), so you get a second level of backslash interpretation. egrep '\' is a syntax error; to grep for a literal single backslash, use egrep '\\', or (like aju_kup suggested, in double quotes, egrep "\\\\" -- each pair of backslashes is reduced by the shell's backslash parsing to a single backslash, so egrep gets two backslashes -- the first one escapes the second, to search for a literal backslash).