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xargs/grep/expr syntax error
Posted on 2008-10-09
I need to execute a command which returns an integer and compare that result to the result of another integer-returning command and get back 0 or 1 if the integers match or not. Here's what I have:
/home/me -T B | grep -c 11 | xargs expr -I{} != /home/me -TC | grep 20081008 | grep -c 11. I'm getting a syntax error. Is what I'm trying to do possible? If so, what's the issue with my syntax?
Thanks in advance,
Nefertiti_IT
/home/me -T B | grep -c 11 | xargs expr -I{} != /home/me -TC | grep 20081008 | grep -c 11. I'm getting a syntax error. Is what I'm trying to do possible? If so, what's the issue with my syntax?
Thanks in advance,
Nefertiti_IT
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7 Comments
#!/bin/bash
res1=`/my/command1`
res2=`/my/command2`
if [ "$res1" = "$res2" ]
then
echo match
else
echo failed
fi
res1=`/my/command1`
res2=`/my/command2`
if [ "$res1" = "$res2" ]
then
echo match
else
echo failed
fi
You can do something like if [ `command1` -eq `command2` ]; then echo match; fi
but you cant do command1|command2 and then compare their return status.
but you cant do command1|command2 and then compare their return status.
to clarify further ... `` would capture the output ... If you mean values returned to shell, you would need $? to get return status of last command
command1
r1=$?
command2
r2=$?
if [ $r1 -eq $r2 ]
then
echo match
else
echo failed
fi
command1
r1=$?
command2
r2=$?
if [ $r1 -eq $r2 ]
then
echo match
else
echo failed
fi
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If you use the $() construct instead of backticks, you get the advantage that you can nest $(). Also I think it's easier to read, but that's only a personal opinion
[ $(/home/me -T B | grep -c 11 | xargs expr -I{}) != $(/home/me -TC | grep 20081008 | grep -c 11) ] && echo 0 || echo 1
Not sure about "expr -I{}" though, what does that do?
[ $(/home/me -T B | grep -c 11 | xargs expr -I{}) != $(/home/me -TC | grep 20081008 | grep -c 11) ] && echo 0 || echo 1
Not sure about "expr -I{}" though, what does that do?
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