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Extract directory from full file name?

 
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# 1  
Old 11-06-2012
Extract directory from full file name?
I think I know what this is doing, but the 'eval' is confusing

Code:
fname=$(echo ${lineItem} | awk 'BEGIN {FS=";"}{print $1}')
fname=${fname%%+([[:space:]])}
fname=${fname##+([[:space:]])}
eval "fname=${fname}"

The first line extracts the contents of the line preceeding the ";"
2nd & 3rd lines trim the value (I think .. not sure what the +([[:space:]]) does ..)
But what would be the purpose of the eval statement?
Last edited by joeyg; 11-07-2012 at 10:46 AM.. Reason: corrected spelling
# 2  
Old 11-06-2012
to get directory of file, try also:
Code:
dname=$(dirname filename)

# 3  
Old 11-06-2012
Actually, I need to know what the original code is doing, as I have to document (pseudo-code) the process. I know some basics, and can typically work out what is being done, but things like this throw me for a loop!

- Jon

PS - The authors may not have been the best coders, and if it is as simple as you indicate, then someone will modify the code!
# 4  
Old 11-07-2012
Since we have no idea what your input looks like and we have no idea what output is being produced, it is hard to guess what the author of this code intended the eval to have the shell expand for you. It may also help to know what shell you're using and what type of system you're using.
# 5  
Old 11-08-2012
The script starts with #!/bin/ksh, hopefully that helps with the shell portion. The contact that can provide the actual server mfg and Unix/Linux version is on vacation for the next few days ... :-(

The input file (aka lineitem) looks like:
/finance/reports/july_operations.gzp;120

So fname starts out as /finance/reports/july_operations.gzp

- Jon
# 6  
Old 11-08-2012
Try to run the script with the -x option set and publish the log.
 
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