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what does "[ $? -eq 0 ]" mean

 
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Top Forums UNIX for Dummies Questions & Answers what does "[ $? -eq 0 ]" mean
# 1  
Old 09-10-2010
what does "[ $? -eq 0 ]" mean
Hi

Can somebody please explain this to me:

Code:
 
[ $? -eq 0 ]

# 2  
Old 09-10-2010
It checks if the return code of the previous command is zero which means everything went well.
# 3  
Old 09-10-2010
Thank you.

I'm trying to also figure out what this script does:

Code:
#!/bin/bash

rc=0

testname=`basename $0`
testfile=/tmp/$testname.dsx
errfile=/tmp/$testname.err

../dsxdiff			\
	--verbose		\
	--compatibility		\
	-otestCell1.dsx		\
	-ntestCell1.dsx		\
	1> $testfile		\
	2> $errfile

grep --silent "failed to emit empty file from testCell1.dsx" $errfile
if [ $? -eq 0 ]; then rc=1; fi

diff testEmpty.dsx $testfile > /dev/null
rc=$[ $rc || $? ]

if [ $rc -eq 0 ]
then
	echo $testname passed
	rm $testfile
	rm $errfile
else
	echo $testname FAILED
fi

exit $rc

# 4  
Old 09-10-2010
  • Defining some variables
  • Calling a script or binary named dsxdiff with various options and writing stdout (1>) in one file and stderr (2>) into another file
  • grep in a file for a pattern and if grep was successful, ie. it's return code was 0, set variable rc=1
  • Getting difference of two files and check the return code of the diff
  • This line looks like an error - does not work on my bash: rc=$[ $rc || $? ]. Looks like intention was to check if $rc is true/set and in case not (||) rc will get the value $?. There is another positional parameter substitution syntax for that one.
  • If rc is 0, remove some files; else tell that the test failed
  • Exit with the return code stored in $rc
This User Gave Thanks to zaxxon For This Post:
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