Kevin Martin wrote:
So I'm fairly confused at this point.
Start with a /bin/sh shell.
Then do:
sh-3.2$ unset $?
sh: unset: `0': not a valid identifier
You mean "unset ?". That's not to say that it'll work, but when you
unset variables, you don't precede them with the $ character.
ok, no problem; then do
sh-3.2$ /bin/ksh -c " set -xv ; grep ABCD b ; echo $? ; if [ "$?" =
"0" ] ; then echo yes ; fi"
+ grep ABCD b
ABCD="C" ; export ABCD
+ echo 1
1
+ [ 1 = 0 ]
WHAT?
If you want to understand what's going on, try:
$ false
$ echo "set -xv ; grep ABCD b ; echo $? ; if [ "$?" ="0" ] ; then echo
yes ; fi"
If you want to pass "$" characters like that, you'll need to escape them
or use single quotes:
$ ksh -c " set -xv ; grep ABCD b ; echo \$? ; if [ \"\$?\" =\"0\" ] ;
then echo yes ; fi"
$ ksh -c ' set -xv ; grep ABCD b ; echo $? ; if [ "$?" ="0" ] ; then
echo yes ; fi'
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