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Bash conditional | getting logic wrong?

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# 1  
Old 03-14-2013
Question Bash conditional | getting logic wrong?
I have a file

Code:
cat <<EOF > /tmp/test
Line one
Line two 
Line three
EOF

Now I am trying to get the exit stat ($?) of 1 if any text is found and 0 if any text is not found using grep, i.e. just reversing the exit status of grep

Code:
# (snippet 1) this one is not working!!! retval $? should be 1
grep -q 'one' /tmp/test && false || true
echo $?


# (snippet 2) but this one is working - retval is 1
grep -q 'one' /tmp/test
if [ $? -eq 0 ]; then 
    false
else 
    true
fi
echo $?

What am I missing in snippet 1?
# 2  
Old 03-14-2013
cmd || true will always return 0, since either cmd succeeds or true does. In your snippet #1, cmd is grep && false. That is always false, since either grep fails or false does.

Your snippet #1 therefore reduces to false || true. That will always return 0.

Since your grep command succeeds:
Code:
(grep && false) || true
(true && false) || true
(false) || true
true

Regards,
Alister
Last edited by alister; 03-14-2013 at 02:45 AM..
This User Gave Thanks to alister For This Post:
the_gripmaster
# 3  
Old 03-14-2013
Is there a way to shorten snippet 2?
# 4  
Old 03-14-2013
Quote:
Originally Posted by the_gripmaster
Is there a way to shorten snippet 2?
Looks like you just want to invert the status:
Code:
! grep -q 'one' /tmp/test

Regards,
Alister
This User Gave Thanks to alister For This Post:
the_gripmaster
# 5  
Old 03-14-2013
Cool. Thanks.
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