new line in bash parameter substitution ${REV%%\n*} - bash

Question

does not work
echo ${REV%%\n*}
does work
echo ${REV%%
*}
After reading trough http://tldp.org/LDP/abs/html/parameter-substitution.html I still can not figure out how to make \n work.

Answer 1

${REV%%$'\n*'} seems to work. See the quoting section of the bash documentation.

Answer 2

If your intent is to try and get it on one line and you're willing to go "outside" of bash, you can use:
echo "$(echo "${REV}" | head -1l)"
But, assuming your version of bash is recent enough, you can try:
pax> export REV="abc
...> def"
pax> echo "${REV}"
abc
def
pax> echo "${REV%%$'\n'*}"
abc
The reason you need $'\n' is because the bash definition of word is somewhat restrictive, compared to what you expect. The bash manpage has this to say:
Words of the form $'string' are treated specially. The word expands
to string, with backslash-escaped characters replaced as specified by
the ANSI C standard. Backslash escape sequences, if present, are decoded
as follows:
\a alert (bell)
\b backspace
\e
\E an escape character
\f form feed
\n new line
\r carriage return
\t horizontal tab
\v vertical tab
\\ backslash
\' single quote
\" double quote
\nnn the eight-bit character whose value is the octal value
nnn (one to three digits)
\xHH the eight-bit character whose value is the hexadecimal
value HH (one or two hex digits)
\uHHHH the Unicode (ISO/IEC 10646) character whose value is
the hexadecimal value HHHH (one to four hex digits)
\UHHHHHHHH
the Unicode (ISO/IEC 10646) character whose value is
the hexadecimal value HHHHHHHH (one to eight hex digits)
\cx a control-x character

Answer 3

More portably (ie, POSIX):
var="yo
yo"
newline="
"
echo "${var#*"$newline"}"