for i in `find . -type f -name "VF-Outlet*.edi" -exec basename \{} \;` ; do
if [ -n "${i}" ];
then
echo file "VF-Outlet found";
sed -e 's/\*UK\*00/\*UP\*/g;s/XQ.*$/XQ\*H\*20150104\*20150110/g' $i > ${i}_fix
else
echo file "VF-Outlet" not found;
fi
done
The above code works if the file is found. The 'echo' statement prints file found. If the file is not found however, nothing prints. I tried all the various tests for empty string, and unset variables, nothing works. Also if I try:
i=`find . -type f -name "VF-Outlet*.edi" -exec basename \{} \;`;
Then do the test:
if [ -n "${i}" ];
then
echo file ${i} found;
else
echo file "VF-Outlet" not found;
fi
done
It works correctly if the file is found or not.
Need help in figuring this out. I need the for loop to test multiple files.
The reason it is not working is due to the fact that "for" does not take null value as input for the variable "i"
For ex:
for i in echo > /dev/null; do echo hi; done
The above command wont give any result, because no value has been assigned to value $i for running the loop.
In the case mentioned here if we check the script in debug mode, we can see that the script dies at initial variable assignment.
# sh -x script.sh + find . -type f -name VF-Outlet*.edi -exec basename {} ;
here, script.sh file contains the script you have provided.
If there is a file present in the directory, the script will successfully execute.
# sh -x tet
+ find . -type f -name VF-Outlet*.edi -exec basename {} ;
+ [ -n VF-Outlet1.edi ]
+ echo file VF-Outlet found
file VF-Outlet found
As @shellter mentioned, this not how I would have done. You can use -f instead of -n to check if a file exists.
Hope this helps!